Find the sum of $\sum_{k=1}^{k=n} \frac{1}{k(k+2)}$.

Question

Here is my solution.

Since $\frac{1}{k(k+2)} = \frac{1}{2} \left(\frac{1}{k} - \frac{1}{k+2}\right)$,

$\sum_{k=1}^{k=n} \frac{1}{k(k+2)} = \frac{1}{2}\sum_{k=1}^{k=n} \left(\frac{1}{k} - \frac{1}{k+2}\right)$

$= \frac{1}{2} \left(\sum_{k=1}^{k=n}\frac{1}{k} - \sum_{k=1}^{k=n}\frac{1}{k+2}\right)$

I am stocked up after this. Please feel free to share your comments and suggestions to help me solve the problem.

Does this answer your question? Formula for the sum $\sum_{i=2}^{n} \frac1{i^2-1}$ — MH.Lee, Oct 31 '21 at 16:01

score 3 · Accepted Answer · answered Oct 31 '21 at 15:55

$\sum_{k=1}^{k=n}\frac{1}{k} - \sum_{k=1}^{k=n}\frac{1}{k+2}=\Big(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n-1}+\frac{1}{n}\Big)-\Big(\frac{1}{3}+\frac{1}{4}+...+\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}\Big)=1+\frac{1}{2}-\frac{1}{n+1}-\frac{1}{n+2}$

score 2 · Answer 2 · answered Oct 31 '21 at 16:25

$$S_n=\sum_{k=1}^n \frac{1}{k(k+2)}=\frac{1}{2} \sum_{k=1}^{n} \left(\frac{1}{k}-\frac{1}{k+2}\right)=\frac{1}{2}\sum_{k=1}^{n}\int_{0}^{1}\left( t^{k-1}-t^{k+1}\right) dt.$$ $$\implies S_n=\frac{1}{2}\int_{0}^{1}(\frac{1}{t}-t)t\frac{t^n-1}{t-1}dt=-\frac{1}{2}\int_{0}^{1}(1+t)(t^n-1) dt$$ $$S_n=-\frac{1}{2}\int_{0}^{1}(t^{n+1}+t^n-t-1) dt=\frac{1}{2}\left(\frac{3}{2}-\frac{1}{n+1}-\frac{1}{n+2}\right).$$ Check that $S_1=\frac{1}{3}.$

score 1 · Answer 3 · answered Oct 31 '21 at 16:35

$$S_n=\sum_{k=1}^n \frac{1}{k(k+2)}=\frac{1}{2} \sum_{k=1}^{n} \left(\frac{1}{k}-\frac{1}{k+2}\right)=\frac{1}{2} \sum_{k=1}^{n}\left[ \left(\frac{1}{k}-\frac{1}{k+1}\right)+\left(\frac{1}{k+1}-\frac{1}{k+2}\right)\right].$$ Here on one can do telescopic summation.

Find the sum of $\sum_{k=1}^{k=n} \frac{1}{k(k+2)}$.

3 Answers3