# The limit of a sum (turning it to a Riemann sum?)

Calculate the limit
$$\lim_{n\to\infty}\sum_{k=1}^{n}\frac{n^2}{(n^2+k^2)(n+k)}$$

I tried solving it by changing it a Riemann sum then integrating, however I couldn't manipulate the algebra to its form. Is there another way of doing this or am I on the right track?

## 1 Answer

HINT:

$\displaystyle\lim_{n\to\infty}\sum_{k=1}^{n}\frac{n^2}{(n^2+k^2)(n+k)}=\lim_{n\to\infty}\frac1n\sum_{k=1}^{n}\frac1{(1+(k/n)^2)(1+k/n)}$

Now, $\displaystyle\lim_{n\to\infty}\frac1n\sum_{k=1}^{n}f(k/n)=\int_0^1f(x)dx$

Please let me know if you can complete the task after using Partial Fraction Decomposition

• Thanks for the hint. Is the integral $\displaystyle\int_0^1 \frac{1}{1+x+x^2+x^3} dx$ ? – codeedoc Mar 16 '14 at 5:07
• @user73645, yes. Then we need to use $$\frac1{1+x+x^2+x^3}=\frac1{(1+x)(1+x^2)}=\frac A{1 +x}+\frac{Bx+C}{1+x^2}$$ – lab bhattacharjee Mar 16 '14 at 5:09
• Oh, how silly I am to expand the denominator! I got it right now! Thank you so much! – codeedoc Mar 16 '14 at 5:14