How to find the sum $\frac{1}{1^4+1^2+1}+\frac{2}{2^4+2^2+1}+..+\frac{2015}{2015^4+2015^2+1}$ ?
I'm not being able to approach the problem.Hints please!
Asked
Active
Viewed 77 times
1
-
2Well, you can cancel $n$ from the top and bottom, to get $\frac1{n^3+n+1}$. – Akiva Weinberger Sep 09 '15 at 14:19
-
I think you'll need the digamma function (as well as the roots of $n^3+n+1$), if you want the general answer. Specifically for 2015, the answer is 0.49472. – Akiva Weinberger Sep 09 '15 at 14:26
-
1I suspect there is a typo and the denominator is $n^4 + n^2 + 1$ instead. – achille hui Sep 09 '15 at 14:27
-
@achillehui oh no!!!!sorry!! – Sep 09 '15 at 14:31
-
3Hint: $\frac{n}{n^4+n^2+1} = \frac{n}{(n^2+1)^2-n^2} = \frac12\left(\frac{1}{n^2-n+1} - \frac{1}{n^2+n+1}\right)$ and... – achille hui Sep 09 '15 at 14:36
-
I sincerely apologize for the typographical error..I typed in a hurry..thanks achille hui for pointing it out :) – Sep 09 '15 at 14:36
-
@achillehui oh yes..done telescopic series :-D good idea! – Sep 09 '15 at 14:36
2 Answers
7
Notice that: $$ \frac{n}{n^4+n^2+1}=\frac{n}{(n^2+n+1)(n^2-n+1)} = \frac{1}{2}\left(\frac{1}{n^2-n+1}-\frac{1}{n^2+n+1}\right)$$ and that: $$ \frac{1}{(n+1)^2-(n+1)+1}=\frac{1}{n^2+n+1}.$$
Together they give: $$\begin{eqnarray*} \sum_{n=1}^{2015}\frac{n}{n^4+n^2+1}&=&\frac{1}{2}\sum_{n=1}^{2015}\left(\frac{1}{n^2-n+1}-\frac{1}{(n+1)^2-(n+1)+1}\right)\\&=&\frac{1}{2}\left(\frac{1}{1}-\frac{1}{2016^2-2016+1}\right)=\color{blue}{\frac{2031120}{4062241}}.\end{eqnarray*}$$
Jack D'Aurizio
- 353,855
0
You might want to prove by induction that the sum
$$\sum_{i=1}^n \frac{i}{i^4+i^2+1} = \frac{\frac{1}{2} n(n+1)}{n^2+n+1}$$
Patrick Stevens
- 36,135