How would I go about computing this finite sum?

Question

How would I go about computing the sum $$ \sum_{k=1}^{n} \dfrac{(-k^2+2k+1)2^k}{(k(k+1))^2}. $$

I have tried partial fractions but have gotten stuck trying to find the coefficients. I decomposed it like this: $$ \dfrac{2^k(-k^2+2k+1)}{(k(k+1))^2} = \frac{a_0}{k} + \frac{a_1}{k^2} + \frac{a_2}{k+1} + \frac{a_3}{(k+1)^2} $$.

Answer 1

Hint:

Without the need for solving the coefficients:

$\dfrac{-k^2+2k+1}{k^2(k+1)^2} =\dfrac{k^2+2k+1-2k^2}{k^2(k+1)^2}=\dfrac{(k+1)^2}{k^2(k+1)^2}-\dfrac{2}{(k+1)^2}=\dfrac{1}{k^2}-\dfrac{2}{(k+1)^2}.$

Thus, the sum is reduced to

$$\sum_{k=1}^{n}\dfrac{2^k}{k^2}-\sum_{k=1}^{n}\dfrac{2^{k+1}}{(k+1)^2}.$$

And then note that

$$\sum_{k=1}^{n}\dfrac{2^{k+1}}{(k+1)^2}=\sum_{k=2}^{n+1}\dfrac{2^{k}}{k^2}=-2+\dfrac{2^{n+1}}{(n+1)^2}+\sum_{k=1}^{n}\dfrac{2^{k}}{k^2}.$$

The rest is for you to finish.

Answer 2

Observe that

$$\dfrac{(-k^2+2k+1)2^k}{(k(k+1))^2}= \dfrac{-2^{k+1}}{(k+1)^2}+\dfrac{2^k}{k^2}.$$

Then take sum as $k$ varies from $1$ to $n$.

$$\sum_{k=1}^n \dfrac{-2^{k+1}}{(k+1)^2}+\dfrac{2^k}{k^2}=\sum_{k=1}^n \dfrac{-2^{k+1}}{(k+1)^2}+\sum_{k=1}^n \dfrac{2^k}{k^2}$$

This being an alternate sum of elements of same type, you are left with $$2-\dfrac{2^{n+1}}{(n+1)^2}$$

Answer 3

The partial fraction expansion gives $$ {{ - k^{\,2} + 2k + 1} \over {\left( {k\left( {k + 1} \right)} \right)^2 }} = {1 \over {k^{\,2} }} - {2 \over {\left( {k + 1} \right)^{\,2} }} $$

Then we have $$ \eqalign{ & {{u^{\,k} } \over k} = \int_{t = 0}^{\,u} {t^{\,k - 1} dt} \quad \Rightarrow \quad {{x^{\,k} } \over {k^{\,2} }} = \int_{u = 0}^x {{{du} \over u}\int_{t = 0}^{\,u} {t^{\,k - 1} dt} } \quad \Rightarrow \cr & \Rightarrow \quad {{x^{\,k} } \over {\left( {k + 1} \right)^{\,2} }} = {1 \over x}\int_{u = 0}^x {{{du} \over u}\int_{t = 0}^{\,u} {t^{\,k} dt} } \cr} $$ so that $$ \eqalign{ & \sum\limits_{k = 1}^n {{{ - k^{\,2} + 2k + 1} \over {\left( {k\left( {k + 1} \right)} \right)^2 }}x^{\,k} } = \sum\limits_{k = 1}^n {{{x^{\,k} } \over {k^{\,2} }}} - 2\sum\limits_{k = 1}^n {{{x^{\,k} } \over {\left( {k + 1} \right)^{\,2} }}} = \cr & = \sum\limits_{k = 1}^n {\int_{u = 0}^x {{{du} \over u}\int_{t = 0}^{\,u} {t^{\,k - 1} dt} } } - 2\sum\limits_{k = 1}^n {{1 \over x}\int_{u = 0}^x {{{du} \over u}\int_{t = 0}^{\,u} {t^{\,k} dt} } } \cr} $$

In particular for $x=2$ we have $$ \eqalign{ & \sum\limits_{k = 1}^n {{{ - k^{\,2} + 2k + 1} \over {\left( {k\left( {k + 1} \right)} \right)^2 }}2^{\,k} } = \sum\limits_{k = 1}^n {\int_{u = 0}^2 {{{du} \over u}\int_{t = 0}^{\,u} {t^{\,k - 1} dt} } } - \sum\limits_{k = 1}^n {\int_{u = 0}^2 {{{du} \over u}\int_{t = 0}^{\,u} {t^{\,k} dt} } } = \cr & = \sum\limits_{k = 1}^n {\int_{u = 0}^2 {{{du} \over u}\int_{t = 0}^{\,u} {t^{\,k - 1} dt} } } - \sum\limits_{k = 1}^n {\int_{u = 0}^2 {{{du} \over u}\int_{t = 0}^{\,u} {t^{\,k} dt} } } = \cr & = \sum\limits_{k = 1}^n {\int_{u = 0}^2 {{{du} \over u}\int_{t = 0}^{\,u} {t^{\,k - 1} \left( {1 - t} \right)dt} } } = \cr & = \cdots \cr} $$

Can you continue from here ?

(the solution is $ 2 - {{2^{n + 1} } \over {\left( {n + 1} \right)^{\,2} }} $ )