Infinite series of formula

Question

How do I compute this infinite series

$$ \sum_{k=0}^{\infty}{\frac{(7n+32)(3^n)}{n(n+2)(4^n)}}$$

I believe partial fraction decomposition is part of the solving method but am a little stuck because the $3^n$ and the $4^n$ make the decomposition method a bit strange:

${\frac{(7n+32)(3^n)}{n(n+2)(4^n)}} = {\frac{A}{n}} + {\frac{B}{n+2}} + {\frac{C}{4^n}}$

$(7n+32)(3^n) = A(n+2)(4^n) + B(n)(4^n) + C(n)(n+2)$

$7n(3^n) + 32(3^n) = An (4^n) + 2A(4^n) + B(n)(4^n) + C(n^2 + 2n)$

$7n(3^n) + 32(3^n) = (An +2A + Bn)(4^n) + C(n^2 + 2n)$

..Stuck on how to proceed.

score 3 · Answer 1 · answered Mar 07 '17 at 06:21

3

HINT:

$$\frac{(7n+32)(3^n)}{n(n+2)(4^n)}=\frac{(4^2(n+2)-3^2)(3^n)}{n(n+2)(4^n)}$$

$$=\dfrac{3^n}{n4^{n-2}}-\dfrac{3^{n+2}}{(n+2)4^n}$$

$$=16\left(\dfrac{(3/4)^n}n-\dfrac{(3/4)^{n+2}}{n+2}\right)$$

answered Mar 07 '17 at 06:21

lab bhattacharjee

279,016

See http://mathworld.wolfram.com/TelescopingSum.html and http://math.stackexchange.com/questions/104918/how-to-analyze-convergence-and-sum-of-a-telescopic-series-i-cant-find-a-generi – lab bhattacharjee Mar 07 '17 at 06:26
And $$\sum_{n=1}^{\infty}\frac{x^n}{n}=-\log(1-x)$$ (+1). – Hazem Orabi Mar 08 '17 at 22:16

Fimpellizzeri · Accepted Answer · 2017-03-07T06:38:08.507

I would do it as follows.

$$ \sum_{n=0}^{\infty}{\frac{(7n+32)(3^n)}{n(n+2)(4^n)}}= 7\sum_{n=0}^{\infty}{\frac{1}{n+2}{\left(\frac34\right)}^n} + 32\sum_{n=0}^{\infty}{\frac{1}{n(n+2)}{\left(\frac34\right)}^n}$$

Now, we have that

$$\sum_{n=0}^{\infty}{\frac{1}{n+2}{\left(\frac34\right)}^n}= {\left(\frac34\right)}^{-2}\cdot\sum_{n=0}^{\infty}{\frac{1}{n+2}{\left(\frac34\right)}^{n+2}}\\ ={\left(\frac34\right)}^{-2}\cdot\sum_{n=2}^{\infty}{\frac{1}{n}{\left(\frac34\right)}^{n}}$$

Now, for $|r|<1$, we have that

$${(1-r)}^{-1}=\sum_{n\geq0}r^n,$$

so that integrating both sides by $r$ yields

$$-\ln(1-r)=\sum_{n\geq0}\frac1{n+1}r^{n+1}=\sum_{n\geq1}\frac1nr^n.$$

It follows that

$$\sum_{n=2}^{\infty}{\frac{1}{n}{\left(\frac34\right)}^{n}}=-\ln\left(1-\frac34\right)-\frac34=2\ln(2)-\frac34.$$

Finally, we use a partial fraction decomposition on $\frac1{n(n+2)}=\frac12\frac1n-\frac12\frac1{n+2}$, and this splits the final series into two series that can be solved like the one we just did. Do you think you can finish it?

EDIT: Well, I just noticed that in your opening question there is a division by $0$ in the $n=0$ term, but this shouldn't change your approach to solving the question. Assuming the summation is carried out for $n\geq1$, you should obtain $\frac{33}2$ in the end.

Infinite series of formula

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