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Because Nick is closer to the Greenwich Meridian he beat me to it!Not fair!OK, that was a joke.It is actually absolutely awesome that we have a number of different solutions of the same problem. A very important part of mathematical culture (of a person) is to do this sort of exercises on a regular basis - throw a problem on the chalkboard and then go at from different angles. The more the merrier.In the chronological order of appearance: in addition to the Gamma function approach by Nick Shales and the partial fraction decomposition approach by Sridhar Ramesh (be sure to study both methods) here are two more angles that I can think of: Binomial Differentials and Telescopicity or Telescopeness or Telescopism or whatever form of Telescope works.3. Binomial Differentials (similar to Nick’s approach)are discussed in this Quora answer which shows that the indefinite integral[math]\displaystyle \int x^m(a [/math][math][/math][math]+ bx^n)^p\; dx \tag*{}[/math]can always be reduced to[math]\displaystyle I_{p,q}(a,b) = \dfrac{1}{n}\cdot\int (a+bx)^p x^q\;dx \tag{1}[/math]From the problem-solving perspective the challenge then is 1) to connect (1) to this particular setting and 2) to discern which choice of [math]a, b, p[/math] and [math]q[/math] for a definite integral will yield the common term of the series requested.Let us see. Let [math]p[/math] and [math]n[/math] be natural numbers. Frame the common term of the generalized series as[math]\dfrac{1}{p\cdot(p+1)\cdot(p+2)\cdot\ldots\cdot(p+n)} \tag*{}[/math]Then the choices [math]a = 1[/math], [math]b = -1[/math], [math]n[/math] in place of [math]p[/math] in (1) and [math](p - 1)[/math] in place of [math]q[/math] work:[math]\displaystyle \dfrac{1}{p\cdot(p+1)\cdot\ldots\cdot(p+n)} = \dfrac{1}{n!}\cdot\int\limits_0^1(1-x)^nx^{p-1}\;dx \tag*{}[/math]Proof: integrating the above RHS by parts once, we reduce the exponent [math]n[/math] by one:[math]\displaystyle \int\limits_0^1(1-x)^nx^{p-1}\;dx = \dfrac{n}{n+p}\cdot\int\limits_0^1(1-x)^{n-1}x^{p-1}\;dx \tag*{}[/math]Repeating the integration by parts until we whittle that exponent down to [math]0[/math], we find:[math]\displaystyle \int\limits_0^1(1-x)^nx^{p-1}\;dx = \dfrac{n!\cdot(p-1)!}{(n+p)!} \tag{2}[/math]Since the factorial [math](n+p)![/math] is compound, we partition it into two portions - in one portion [math]n[/math] is fixed at [math]0[/math] and the constituent integers vary from [math]1[/math] to [math]p[/math] amounting to [math]p![/math] while in the other portion [math]p[/math] is fixed and the constituent integers vary from [math]1[/math] to [math]n[/math] amounting to [math](p+1)\cdot\ldots\cdot(p+n)[/math].Hence, cancelling the common [math]([/math][math]p-1[/math][math])![/math] factor in both numerator and denominator in (2), we find:[math]\dfrac{n!\cdot(p-1)!}{(n+p)!} = \dfrac{n!}{p\cdot(p+1)\cdot\ldots\cdot(p+n)} \tag*{}[/math]which concludes the proof.In your case [math]n = 3[/math] and for any natural number [math]p[/math] (which will vary) the common term of the series is equal to[math]\dfrac{1}{p\cdot(p+1)\cdot(p+2)\cdot(p+3)} \tag*{}[/math]Thus, the remaining steps, which employ the geometric series interim, are mechanical:[math]\displaystyle \sigma = \sum_{p=1}^{+\infty}\dfrac{1}{p\cdot(p+1)\cdot(p+2)\cdot(p+3)} = \sum_{p=1}^{+\infty}\dfrac{1}{3!}\cdot\int\limits_0^1(1-x)^3x^{p-1}\;dx \tag*{}[/math]After justifying the right to move the summation sign through the integral sign, we have:[math]\displaystyle \sigma = \dfrac{1}{3!}\cdot\int\limits_0^1(1-x)^3\sum_{p=1}^{+\infty}x^{p-1}\;dx = \tag*{}[/math][math]\displaystyle \dfrac{1}{3!}\cdot\int\limits_0^1(1-x)^3\cdot\dfrac{1}{1-x}\;dx = \dfrac{1}{3!}\cdot\int\limits_0^1(1-x)^2\;dx = \tag*{}[/math][math]-\;\dfrac{1}{3!}\cdot\dfrac{(1-x)^3}{3}\Bigg|_0^1 = \dfrac{1}{3!}\cdot\dfrac{1}{3} = \dfrac{1}{18} \tag*{}[/math]which is to say that[math]\displaystyle \sum_{p=1}^{+\infty}\dfrac{1}{p\cdot(p+1)\cdot(p+2)\cdot(p+3)} = \dfrac{1}{18} \tag*{}[/math]4. Telescopicity (similar to Sridhar’s approach)If all the terms [math]A_i[/math] of a (finite) sum [math]S_n[/math] such that[math]S_n = A_1 [/math][math][/math][math]+ A_2 [/math][math][/math][math]+ A_3 [/math][math][/math][math]+ \ldots [/math][math][/math][math]+ A_n \tag*{}[/math]can, through some transformation, be put into a form which amounts to a difference between two adjacent terms [math]a_i[/math] and [math]a_{i-1}[/math] in some order as, for one but not the only example, in [math]A_1 = a_1 - a_0[/math] and [math]A_2 = a_2 - a_1[/math] and [math]A_3 = a_3 - a_2[/math] and so on, then[math]S_n = a_1 - a_0 [/math][math][/math][math]+ a_2 - a_1 [/math][math][/math][math]+ a_3 - a_2 [/math][math][/math][math]+ \tag*{}[/math][math]\ldots [/math][math][/math][math]+ a_n - a_{n-1} = a_n - a_0 \tag*{}[/math]and we see that all the intermediate terms of [math]S_n[/math] cancel out and only the first and the last terms survive. We thus obtain for [math]S_n[/math] something that we frequently refer to but never rigorously define, to the best of my knowledge, as closed form.In any case, if the above transformation is possible then the sum [math]S_n[/math] is said to be telescopic.The challenge and the fun part now is to see if we can expose or lay bare the telescopic nature of the sum requested.In the upcoming deluge of symbols shuffling it pays to keep the following motivational toy model, which we describe as our game plan, front and center.Let the [math]n-[/math]th common term of a series be [math]A_n[/math]:[math]A_n = \dfrac{1}{2\cdot 3\cdot 4\cdot 5} \tag{3}[/math]An idea: instead of fighting the reciprocal products let us fight the direct products of integers by flipping the [math]A_n[/math] in (3) up side down:[math]\dfrac{1}{A_n} = 2\cdot 3\cdot 4\cdot 5 \tag{4}[/math]Then, by shifting the index [math]n[/math] in (4) forward once, that is by replacing [math]n[/math] with [math](n+1)[/math], we will get:[math]\dfrac{1}{A_{n+1}} = 3\cdot 4\cdot 5\cdot 6 \tag{5}[/math]Notice that between them (4) and (5) have a certain commonality which we have highlighted in the diagram below:Can that commonality be exploited?Hm.From (4) it follows that[math]1 = A_n\cdot 2\cdot 3\cdot 4\cdot 5 \tag{6}[/math]Likewise, from (5) we have[math]1 = A_{n+1}\cdot 3\cdot 4\cdot 5\cdot 6 \tag{7}[/math]But unity is one and the same. Thus, the LHSs of (6) and (7) must also be equal:[math]A_n\cdot 2\cdot 3\cdot 4\cdot 5 [/math][math][/math][math] = A_{n+1}\cdot 3\cdot 4\cdot 5\cdot 6 \tag*{}[/math]Observe the following pattern that we will soon reuse: all the integers but the first one of [math]A_n[/math] and all the integers but the last one of [math]A_{n+1}[/math] will cancel out. Make a mental note of that fact as an *:[math]A_n\cdot 2 = A_{n+1}\cdot 6 \tag{8}[/math]In exactly the same way, by shifting the index [math]n[/math] in (4) backward once, that is by replacing [math]n[/math] with [math](n-1)[/math], we will get:[math]\dfrac{1}{A_{n-1}} = 1\cdot 2\cdot 3\cdot 4 \tag*{}[/math]from where we see that the same pattern * holds:and from where[math]A_{n-1}\cdot 1 = A_n\cdot 5 \tag{9}[/math]Lastly, we will name the LHS of (8) as [math]a_{n+1}[/math], the LHS of (9) - as [math]a_n[/math] and then we will construct their difference to see where it takes us.Now for the perfect storm of symbols - we generalize the bejeezuz out of the integers that make up our toy terms as follows: the concrete [math]1\cdot 2\cdot 3\cdot 4[/math] and such become the generic[math]\dfrac{1}{A_n}=(p+nq)(p+(n+1)q)(p+(n+2)q)\ldots(p+(n+m-1)q) \tag{10}[/math]which contains exactly [math]m[/math] parenthesised factors.The integers [math]p, q[/math] and [math]m[/math] are the constants dictated by the particular structure of the given terms and the only varying quantity in (10) is [math]n[/math].In your case, for example, [math]p=0, q=1[/math] and [math]m=4[/math].In (10) replace [math]n[/math] with [math](n+1)[/math]:[math]\dfrac{1}{A_{n+1}}=(p+(n+1)q)(p+(n+2)q)(p+(n+3)q)\ldots(p+(n+m)q) \tag{11}[/math]The commonality between (10) and (11), as we explained above*, is:from where:[math]A_n\cdot(p+nq) = A_{n+1}\cdot(p+(n+m)q) \tag*{}[/math]and we name the result as [math]a_{n+1}[/math]:[math]a_{n+1} = A_{n+1}\cdot(p+(n+m)q) \tag{12}[/math]In exactly the same way, in (10) replace [math]n[/math] with [math](n-1)[/math]:[math]\dfrac{1}{A_{n-1}}=(p+(n-1)q)(p+nq)(p+(n+1)q)\ldots(p+(n+m-2)q) \tag{11}[/math]to see that the same pattern * holds:and[math]A_{n-1}\cdot(p+(n-1)q) = A_n\cdot(p+(n+m-1)q) \tag*{}[/math]Name the result as [math]a_n[/math]:[math]a_n = A_n\cdot(p+(n+m-1)q) \tag{13}[/math]Subtract (12) from (13):[math]a_n - a_{n+1} = A_n\cdot(m-1)q \tag{14}[/math]noting that in (14) the difference does not explicitly depend on [math]n[/math] but only on [math]A_n[/math] and the rest of the objects are constant numbers. This is what our telescopic recipe prescribes.In (14) replace [math]n[/math] with [math](n-1)[/math]:[math]a_{n-1} - a_n = A_{n-1}\cdot(m-1)q \tag*{}[/math]Repeat:[math]a_{n-2} - a_{n-1} = A_{n-2}\cdot(m-1)q \tag*{}[/math]again and again until [math]A_1[/math] (or [math]A_0[/math] as the case may be) is reached:[math]a_1 - a_2 = A_1\cdot(m-1)q \tag*{}[/math]Add all these differences up: their LHS will telescope into just two terms and their RHS will amount to the partial sum [math]S_n[/math]:[math]a_1 - a_n = S_n\cdot(m-1)q \tag*{}[/math]from where:[math]S_n = \dfrac{a_1 - a_n}{(m-1)q} \tag*{}[/math]and, taking the [math]a_1[/math] and [math]a_n[/math] from (13), we have:[math]S_n = \dfrac{A_1\cdot(p+mq) - A_n\cdot(p+nq)}{(m-1)q} \tag{15}[/math]In a concrete sum the first term in (15)’s numerator will be just a number and the denominator in (15) will be just a number. As such, we might just as well agree on the following Konstant:[math]K = A_1\cdot\dfrac{p+mq}{(m-1)q} \tag{16}[/math]and then[math]S_n = K - A_n\cdot\dfrac{p+nq}{(m-1)q} \tag*{}[/math]If we replace [math]A_n[/math] with its image from (10) and note that the factor [math](p+nq)[/math] common to both the numerator and the denominator in (15) cancels out then the closed form for the generic partial sums in question comes out to be:[math]S_n = K - \dfrac{1}{(m-1)q}\cdot\dfrac{1}{(p+(n+1)q)\ldots(p+(n+m-1)q)} \tag{17}[/math]In your case, as we have mentioned before, [math]p = 0, q = 1, m = 4[/math]. Hence, due to (16):[math]K = \dfrac{1}{1\cdot 2\cdot 3\cdot 4}\cdot\dfrac{0+4\cdot 1}{(4-1)\cdot 1} = \dfrac{1}{18} \tag*{}[/math]and due to (17):[math]S_n = \dfrac{1}{18} - \dfrac{1}{3(n+1)(n+2)(n+3)} \tag*{}[/math]Taking the limit of the above sequence of partial sums as [math]n[/math] tends to [math]+\infty[/math], we arrive at the answer:[math]\displaystyle \sum_{n=1}^{+\infty}\dfrac{1}{n(n+1)(n+2)(n+3)} = \dfrac{1}{18} \tag*{}[/math]

We seek the value of the following limit [math]L_p[/math]:[math]\displaystyle L_p = \lim_{n\to+\infty}\dfrac{1^p+2^p+\ldots +n^p}{n^{p+1}} \tag*{}[/math]Whoever constructed this particular limit went out of his/her way to camouflage the fact that this is actually a good old Riemann integral in disguise. Remember that the classical essence of a Riemann integral is the area of a planar region trapped under a certain curve.We choose a Cartesian [math]xOy[/math] coordinate system, a [math][0, 1][/math] interval on the [math]x[/math]-axis and divide it into [math]n[/math] smaller line segments of equal length [math]\Delta x[/math] tight - with no gaps and no overlaps:[math]\Delta x = \dfrac{1}{n}, [/math][math][/math][math]\; n \in \mathbb{N} \tag{1}[/math]Then, the address or the position of our [math]k[/math]-th primitive line segment on the [math]x[/math]-axis is just [math]k[/math] units away from the origin:[math]x_k = \dfrac{k}{n} \tag{2}[/math]The magnitude of a function [math]f(x)[/math] at the point [math]x_k[/math] is, then, [math]f(x_k)[/math] and together with [math]x_k[/math] these two magnitudes define a primitive rectangle whose width is [math]\Delta x[/math], whose height is [math]f(x_k)[/math] and whose area [math]A_k[/math] is taken to be known a priori:[math]A_k = f(x_k) \cdot \Delta x = f\left(\dfrac{k}{n}\right)\cdot \dfrac{1}{n} \tag*{}[/math]The aggregate sum [math]A'[/math] of the areas of all the primitive rectangles, in general, is an approximation of the area of the planar region in question:[math]\displaystyle A' = \sum_{k=1}^n A_k \tag{3}[/math]and the actual area of the region in question, we claim, is the limit of (3) as [math]n[/math] tends to positive infinity:[math]\displaystyle A = \lim_{n \to [/math][math][/math][math]+ \infty} \sum_{k=1}^n A_k \tag{4}[/math]or:[math]\displaystyle A = \lim_{n \to [/math][math][/math][math]+ \infty} \dfrac{1}{n}\sum_{k=1}^n f\left(\dfrac{k}{n}\right) \tag{5}[/math]From the art of problem solving perspective, if we can find the shape of the mystery function [math]f(k/n)[/math] hiding in the given limit then we can start seeing the light at the end of the tunnel.In this particular case the numbers [math]1^p, 2^p, 3^p[/math], etc. can be folded into a uniform [math]k^p[/math] shape, which can be paired up with [math]n^p[/math] from the denominator as follows:[math]\left(\dfrac{k}{n}\right)^p \tag*{}[/math]and the shape of the said function [math]f(x)[/math], magically, reveals itself:[math]f(x) = x^p \tag*{}[/math]where the nature of the number [math]p[/math] is not clarified in the problem statement, so let us take it to be a natural number, [math]p>0[/math].Also notice that after the above partition of the given summands, we will have exactly one copy of [math]1/n[/math] left over.As such, the requested limit [math]L_p[/math] can rewritten differently but equivalently:[math]\displaystyle L_p = \lim_{n \to [/math][math][/math][math]+ \infty} \dfrac{1}{n}\sum_{k=1}^n \left(\dfrac{k}{n}\right)^p \tag*{}[/math]or:[math]\displaystyle \lim_{n \to [/math][math][/math][math]+ \infty} \dfrac{1}{n^{p+1}}\sum_{k=1}^n k^p \tag*{}[/math]Thus, our ability to come up with the answer to this question hinges on our ability to compute the following sum:[math]\displaystyle \sum_{k=1}^n k^p \tag{6}[/math]In this Quora answer, we already demonstrated a simple but laborious method of computing such sums where [math]p \in \mathbb{N}[/math] with just the basic, high school, knowledge of algebra.The main laborious aspect of the method for computing the sums of type (6) is its strict sequentiality. In order to compute the sum of squares of consecutive natural numbers, we need to compute the sum of the corresponding first powers first. In order to compute the sum of cubes of consecutive natural numbers, we need to compute the sum of squares first and so on. In order to compute the sum of the [math]p[/math]-th powers of consecutive natural numbers, we must first compute the sum of the [math]([/math][math]p-1[/math][math])[/math]-st powers of the said numbers.Using this method it can be shown, and this is a good exercise of its own, that the sum of type (6) can always be represented as follows:[math]\dfrac{n^{p+1}}{p+1} [/math][math][/math][math]+ An^p [/math][math][/math][math]+ Bn^{p-1} [/math][math][/math][math]+ Cn^{p-2} [/math][math][/math][math]+ \dots \tag{7}[/math]for some constant real numbers [math]A, B, C[/math], etc.Now, if you like beautiful mathematics then look up the Bernoulli numbers [math]B_k[/math] - with their help the sum (6) can be written in closed form as follows:[math]\displaystyle \sum_{k=1}^n k^p = \dfrac{n^{p+1}}{p+1} [/math][math][/math][math]+ \sum_{k=1}^{p+1} \dfrac{B_k}{k}\binom{p}{k-1}n^{p-k+1} \tag*{}[/math](see the Faulhaber's formula for more)Coming back to our limit: we see that as [math]n[/math] tends to positive infinity, all the terms in the sum (7) but the first one can be safely thrown out and our limit collapses into:[math]\displaystyle L_p=\lim_{n \to [/math][math][/math][math]+ \infty} \dfrac{1}{n^{p+1}} \cdot \dfrac{n^{p+1}}{p+1} = \dfrac{1}{p+1} \tag*{}[/math]which is the answer:[math]\displaystyle L_p = \lim_{n \to [/math][math][/math][math]+ \infty}\dfrac {1^p +2^p +\ldots +n^p}{n^{p+1}} = \dfrac{1}{p+1} \tag*{}[/math]For example, for [math]p = 1[/math] we find:[math]\displaystyle \sum_{k=1}^n k = \dfrac{n(n+1)}{2} \tag*{}[/math]and:[math]\displaystyle L_1=\lim_{n \to [/math][math][/math][math]+ \infty} \dfrac{1}{n^2} \cdot \dfrac{n(n+1)}{2} = \lim_{n \to [/math][math][/math][math]+ \infty}\left(\dfrac{1}{2} [/math][math][/math][math]+ \dfrac{1}{2n}\right) = \dfrac{1}{2} \tag*{}[/math]You are encouraged to compute the answers for the next few values of [math]p = 2, 3[/math] and, say, [math]4[/math] by hand.