If $k > 1$, then $\frac1{(k - 1)^2}-\frac1{(k + 1)^2}=\frac{4k}{(k^2 - 1)^2}$, hence simplify $\sum\limits_{k = 2}^n\frac k{(k^2 - 1)^2}$

Question

I have the following problem:

Prove that if $k > 1$, then

$$\dfrac{1}{(k - 1)^2} - \dfrac{1}{(k + 1)^2} = \dfrac{4k}{(k^2 - 1)^2}$$

Hence simplify

$$\sum\limits_{k = 2}^n \dfrac{k}{(k^2 - 1)^2}$$

My Work:

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

I could have continued and multiplied the denominator out, but, at this point, I'm thinking that I'm actually expected to have proceeded differently, rather than multiplying the denominator out fully? And I don't have solutions for this problem, so I can't check anything.

I would appreciate it if someone could please take the time to explain this problem.

Answer 1

Here are some guidelines:

• Expand $(k+1)^2-(k-1)^2$, or use the rule $a^2-b^2=(a-b)(a+b)$.

• Notice that $(k-1)(k+1)=k^2-1$, thus $(k-1)^2(k+1)^2=?$

• Once you show that $$\dfrac{1}{(k - 1)^2} - \dfrac{1}{(k + 1)^2} = \dfrac{4k}{(k^2 - 1)^2},$$ you can use it so that the sum becomes\begin{align*} \sum_{k=2}^n\dfrac{k}{(k^2-1)^2}&=\dfrac{1}{4}\sum_{k=2}^n\left(\dfrac{1}{(k-1)^2}-\dfrac{1}{(k+1)^2}\right)\\ &=\dfrac{1}{4}\left(\sum_{k=2}^n\dfrac{1}{(k-1)^2}-\sum_{k=2}^n\frac{1}{(k+1)^2}\right); \end{align*} notice that by changing the variable $k-1$ into $k$, you get $$\sum_{k=2}^n\dfrac{1}{(k-1)^2}=\sum_{k=1}^{n-1}\dfrac{1}{k^2}.$$ Do the same for $\sum_{k=2}^n\frac{1}{(k+1)^2}$ and see what you'll get.

Answer 2

Your approach is right.$$\text{Numerator}=(k+1)^2-(k-1)^2=k^2+2k+1-(k^2-2k+1)=4k\\\text{Denominator}=(k+1)^2(k-1)^2=(k^2-1)^2$$For the second part use telescopic rule. Can you finish now?