1
$\begingroup$

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: $$ \dfrac{2^k(-k^2+2k+1)}{(k(k+1))^2} = \frac{a_0}{k} + \frac{a_1}{k^2} + \frac{a_2}{k+1} + \frac{a_3}{(k+1)^2} $$.

$\endgroup$
  • 3
    $\begingroup$ What have you tried? Have you tried partial fractions or simply rewriting the numerator? $\endgroup$ – Simply Beautiful Art Sep 2 at 14:34
  • 1
    $\begingroup$ Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close without reading the comments. $\endgroup$ – saulspatz Sep 2 at 14:40
  • $\begingroup$ What make you think there is a "closed formula". It is very likely that no such formula exists. As suggested by Simply Beautiful Art, you could at least try by re-writing the numerator under the form $-(k^2+k)+(3k+1)$ $\endgroup$ – Jean Marie Sep 2 at 14:45
  • $\begingroup$ @SimplyBeautifulArt see edit. $\endgroup$ – Lukas Zamora Sep 2 at 14:45
  • 2
    $\begingroup$ You don't want the $2^k$ in your partial fractions. First do the partial fractions of $\frac{-k^2+2k+1}{[k(k+1)]^2}$, then multiply the whole thing by $2^k$. $\endgroup$ – Simply Beautiful Art Sep 2 at 14:48
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ The denominator is $[k(k+1)]^2$, not $k(k+1)^2$. $\endgroup$ – Simply Beautiful Art Sep 2 at 16:26
  • $\begingroup$ Thanks for noting, that will change alot.. $\endgroup$ – M.P Sep 2 at 16:28
0
$\begingroup$

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

$\endgroup$
  • 2
    $\begingroup$ In my honest opinion, this should be a comment. There's no explanation, and there's no answer in this "answer". $\endgroup$ – Simply Beautiful Art Sep 2 at 15:14
  • $\begingroup$ @SimplyBeautifulArt This is a hint... $\endgroup$ – Praphulla Koushik Sep 2 at 15:51
  • 1
    $\begingroup$ Being a hint is not excuse for a poor answer. $\endgroup$ – Simply Beautiful Art Sep 2 at 15:55
  • $\begingroup$ @SimplyBeautifulArt I can’t convince you.. upto you to think what ever you want to think $\endgroup$ – Praphulla Koushik Sep 2 at 15:58
0
$\begingroup$

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} }} $ )

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.