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$$\sum_{n=1}^{\infty}\frac{n^{2}}{\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n+4\right)}$$

MY Approach$\sum_{n=1}^{\infty}$$\frac{n^{2}}{\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n+4\right)}$ = Lim$_{k\rightarrow\infty}$$\sum_{k=1}^{n}\frac{k^{2}}{\left(k+1\right)\left(k+2\right)\left(k+3\right)\left(k+4\right)}$

$\frac{k^{2}}{\left(k+1\right)\left(k+2\right)\left(k+3\right)\left(k+4\right)}$= $\frac{1}{6\left(n+1\right)}$-$\frac{2}{\left(n+2\right)}$+$\frac{9}{\left(n+3\right)2}$-$\frac{8}{\left(n+4\right)3}$

I don't think i can telescope it, And i don't know any other method

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    $\begingroup$ I think you can telescope it. I don't think your partial fractions are correct though. $\endgroup$ – Angina Seng Jan 7 '18 at 17:11
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    $\begingroup$ @LordSharktheUnknown I agree. The fact that there is no term in $n^3$ in the numerator of the fraction means that the coefficients of the partial fractions should add to zero. It will require four terms to be gathered together to make this work. [If there were $n^3$ in the numerator the original fractions would all be like $\frac 1n$ and divergence would be expected - so no telescoping]. $\endgroup$ – Mark Bennet Jan 7 '18 at 17:13
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    $\begingroup$ A bit of a hammer, but this could also work: $$ \sum_{i=1}^\infty\frac{i^2}{(i+1)(i+2)(i+3)(i+4)} = \int_0^1\int_0^z\int_0^y\int_0^x\left(t\frac{\text{d}}{\text{d}t}\left(t\frac{\text{d}}{\text{d}t}\frac1{1-t}\right)\right)\text{d}t\text{d}x\text{d}y\text{d}z $$ $\endgroup$ – user3002473 Jan 7 '18 at 17:15
  • $\begingroup$ OP's partial fraction decomposition is almost correct, but not quite. wolframalpha.com/input/… $\endgroup$ – Alfred Yerger Jan 7 '18 at 17:21
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This is one way to rewrite the summand into a form suitable for telescoping by hand.

Let $p_k(n) = \prod_{\ell=1}^k (n+\ell)$. The series at hand has the form

$$\mathcal{S} \stackrel{def}{=} \sum_{n=1}^\infty \frac{n^2}{(n+1)(n+2)(n+3)(n+4)} = \sum_{n=1}^\infty \frac{n^2}{p_4(n)}$$

Notice we have the identity $$\frac{1}{p_{k+1}(n)} = \frac{1}{k} \frac{(n+k+1)-(n+1)}{p_{k+1}(n)} = \frac{1}{kp_k(n)} - \frac{1}{kp_k(n+1)}$$

Repeat apply this identity to the summand of above series, we find

$$\begin{align} \frac{n^2}{p_4(n)} &= \frac{n^2}{3p_3(n)} - \frac{n^2}{3p_3(n+1)} = \color{red}{\frac{2n-1}{3p_3(n)}} + \color{green}{\frac{(n-1)^2}{3p_3(n)}} - \frac{n^2}{3p_3(n+1)}\\ \color{red}{\frac{2n-1}{3p_3(n)}} &= \frac{2n-1}{6p_2(n)} - \frac{2n-1}{6p_2(n+1)} = \color{orange}{\frac{1}{3p_2(n)}} + \color{blue}{\frac{2n-3}{6p_2(n)}} - \frac{2n-1}{6p_2(n+1)}\\ \color{orange}{\frac{1}{3p_2(n)}} &= \color{magenta}{\frac{1}{3p_1(n)}} - \frac{1}{3p_1(n+1)} \end{align}$$ Combine these, we can rewrite the summand as $\displaystyle\;\frac{n^2}{p_4(n)} = g(n) - g(n+1)$ where $$\begin{align} g(n) &=\color{magenta}{\frac{1}{3p_1(n)}} + \color{blue}{\frac{2n-3}{6p_2(n)}} + \color{green}{\frac{(n-1)^2}{3p_3(n)}}\\ &= \frac{\color{magenta}{2(n+2)(n+3)} + \color{blue}{(2n-3)(n+3)} + \color{green}{2(n-1)^2}}{6(n+1)(n+2)(n+3)}\\ &= \frac{6n^2+9n+5}{6(n+1)(n+2)(n+3)} \end{align}$$

The series is now a telescoping one and $$\mathcal{S} = \lim_{p\to\infty} \sum_{n=1}^p \frac{n^2}{p_4(n)} = \lim_{p\to\infty}( g(1) - g(p+1)) = g(1) = \frac{6+9+5}{6\cdot 2\cdot 3 \cdot 4} = \frac{5}{36}$$

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  • $\begingroup$ Sir,it will be very helpful if you suggest a reference textbook containing these telescopic methods or any related material for learning $\endgroup$ – Mohan Sharma Jan 8 '18 at 7:14
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    $\begingroup$ @MohanSharma I didn't learn this methods from any textbook. it is a result of staying on math.SE long enough and seeing enough questions/answers. After a while, you will notice nothing is new under the sun and similar questions keep on showing up... $\endgroup$ – achille hui Jan 8 '18 at 8:55
  • $\begingroup$ Sir Why did you break $\frac{n^{2}}{3P_{3}\left(n\right)}$ into $\frac{2n-1}{3P_{3}\left(n\right)}$+$\frac{\left(n-1\right)^{2}}{3P_{3}\left(n\right)}$ Sir,How did you know that you need $\frac{2n-1}{3P_{3}\left(n\right)}$.Why did you choose to break $\frac{2n-1}{3P_{3}\left(n\right)}$ ,not $\frac{\left(n-1\right)^{2}}{3P_{3}\left(n\right)}$? $\endgroup$ – Mohan Sharma Jan 8 '18 at 15:03
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    $\begingroup$ @MohanSharma we want to turn the summand into a form suitable for telescoping. The $2^{nd}$ piece $\frac{(n-1)^2}{3P_3(n)}$ already forms a telescoping pair with the $3^{rd}$ piece $\frac{n^2}{3P_3(n+1)}$. You definitely want to keep it. So we work on the $1^{st}$ piece instead. Since it has a lower degree than the rest, if one repeat this procedure, the procedure will terminate in finitely many steps... $\endgroup$ – achille hui Jan 8 '18 at 15:08
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$$\sum_{n \geq 1} \frac{(n!)(n^2)}{(n+4)!}$$


Note,

$$a_n=\frac{1}{6} \frac{(n!)(3!)}{(n+4)!}=\frac{1}{6}\frac{\Gamma(n+1) \Gamma (4)}{\Gamma (n+5)}$$

$$=\frac{1}{6}B(n+1,4)$$

$$=\frac{1}{6} \int_{0}^{1}x^n (1-x)^3 dx$$

Where $\Gamma$ denotes the Gamma Function, and $B$ denotes the Beta Function. We have utilized the Beta-Gamma Function relationship.


We are interested in:

$$\sum_{n \geq 1} n^2a_n$$

$$=\frac{1}{6} \int_{0}^{1} \left(\sum_{n \geq 1} n^2 x^n \right) (1-x)^3 dx$$


It is standard to show,

$$\sum_{n \geq 1} n^2 x^n=\frac{x(x+1)}{(1-x)^3}$$

By first considering $\sum_{n \geq 1} x^n=\frac{1}{1-x}$ and (repeatedly) differentiating then multiplying by $x$ on both sides. Convergence of the above series is for $|x|<1$ by the ratio test.


Hence we get,

$$=\frac{1}{6} \int_{0}^{1} x(x+1) dx$$

$$=\frac{5}{36}$$

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  • $\begingroup$ Sir,it will be very helpful if you suggest a reference textbook containing these telescopic methods or any related material for learning $\endgroup$ – Mohan Sharma Jan 8 '18 at 7:15
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The partial fractions should come out to be $$\frac 16\cdot\frac 1{r+1}-2\cdot\frac 1{r+2}+\frac 92\cdot \frac 1{r+3}-\frac 83\cdot\frac 1{r+4}$$

Examine now what happens to the fractions with denominator $5$, which comes with $r=1,2,3,4$ and you get $$\frac 15\cdot\left(-\frac 83+\frac 92-2+\frac 16\right)=0$$So you do get telescoping. I'll leave you to fill in the remaining details.

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  • $\begingroup$ Sir,it will be very helpful if you suggest a reference textbook containing these telescopic methods or any related material for learning $\endgroup$ – Mohan Sharma Jan 8 '18 at 7:14
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Another approach , applicable to this type of sums in general, is to decompose the fraction as a sum of fractions with Rising Factorials at the denominator. $$ \eqalign{ & {{n^{\,2} } \over {\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)\left( {n + 4} \right)}} = {{\left( {n + 1} \right)\left( {n + 2} \right) - 3n - 2} \over {\left( {n + 1} \right)^{\,\overline {\,4\,} } }} = \cr & = {{\left( {n + 1} \right)^{\,\overline {\,2\,} } - 3\left( {n + 1} \right) + 1} \over {\left( {n + 1} \right)^{\,\overline {\,4\,} } }} = {1 \over {\left( {n + 3} \right)^{\,\overline {\,2\,} } }} - 3{1 \over {\left( {n + 2} \right)^{\,\overline {\,3\,} } }} + {1 \over {\left( {n + 1} \right)^{\,\overline {\,4\,} } }} \cr} $$

That is of advantage because:
- a polynomial at the numerator (here the $n^2$) can always be expressed (through Stirling Numbers) into a "polynomial" in rising/falling "powers";
- the sum of Falling factorials parallels the integral of normal powers;
- you can easily express the sum also between finite bounds.

We have in fact this equivalence between Falling and Rising Factorials $$ x^{\underline {\, - q\,} } = {1 \over {\left( {x + q} \right)^{\underline {\,q\,} } }} = {1 \over {\left( {x + 1} \right)^{\overline {\,q\,} } }} $$ and the indefinite summation of the Rising Factorial is $$ \sum\nolimits_x {{1 \over {\left( {x + 1} \right)^{\overline {\,q\,} } }}} = \sum\nolimits_x {x^{\underline {\, - q\,} } } = {1 \over {1 - q}}x^{\underline {\,1 - q\,} } + c = - {1 \over {\left( {q - 1} \right)\left( {x + 1} \right)^{\overline {\,q - 1\,} } }} + c $$ as it easy to check by taking the Forward Finite Difference of both sides.

Therefore $$ \sum\limits_{n = 1}^\infty {{1 \over {\left( {n + m} \right)^{\overline {\,q\,} } }}} \quad \left| {\;2 \le q} \right.\quad = {1 \over {\left( {q - 1} \right)\left( {m + 1} \right)^{\overline {\,q - 1\,} } }} $$

and the result is $$ \eqalign{ & \sum\limits_{n = 1}^\infty {{{n^{\,2} } \over {\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)\left( {n + 4} \right)}}} = \cr & = \sum\limits_{n = 1}^\infty {\left( {{1 \over {\left( {n + 3} \right)^{\,\overline {\,2\,} } }} - 3{1 \over {\left( {n + 2} \right)^{\,\overline {\,3\,} } }} + {1 \over {\left( {n + 1} \right)^{\,\overline {\,4\,} } }}} \right)} \cr & = {1 \over {1\left( {3 + 1} \right)^{\overline {\,1\,} } }} - {3 \over {2\left( {2 + 1} \right)^{\overline {\,2\,} } }} + {1 \over {3\left( {1 + 1} \right)^{\overline {\,3\,} } }} = \cr & = {1 \over 4} - {1 \over 8} + {1 \over {72}} = {{5} \over {36}} \cr} $$

For more details, also refer to this and to this other related post.

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  • $\begingroup$ Sir,it will be very helpful if you suggest a reference textbook containing these telescopic methods or any related material for learning $\endgroup$ – Mohan Sharma Jan 8 '18 at 7:15
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    $\begingroup$ @MohanSharma: Falling Factorials are a fundamental chapter of Discrete Mathematics, and any good textbook on the subject will treat them. However, I found the renowned "Concrete Mathematics", insuperable, also because it contains a clean and clear exposition of summation and indefinite summation. $\endgroup$ – G Cab Jan 8 '18 at 18:17

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