# where the given series converge to?

Let $$S_n = \sum_{k=1}^{n} \frac{(k^2+nk)}{n^3}$$ then the sequence $$s_n$$

(a) converges to $$\frac{2}{3}$$

(b) Converges to $$\frac{5}{6}$$

(c) converges to $$\frac{6}{5}$$

(d) does not converge

To know where the sequence is converging, I will be required to obtain the where the series $$\sum_{k=1}^{\infty} \frac{(k^2+nk)}{n^3}$$ is converging.

If I write it as $$k^2\sum \frac{1}{n^3}+k\frac{1}{n^2}$$, I know both the series are convergent so the sum is also convergent but where..?

I have dealed with the sums checking whether the series is convergent or not... And also obtaining the sum of geometric series...

How to deal with this sum...no idea!

• You cannot write it as $$k^2\sum_{k=1}^n \left( \frac{1}{n^3}+k\frac{1}{n^2} \right)$$ because use of symbol $k$ doesn't make sense outside the sum.
– Stef
Nov 21 at 2:13

$$\sum_{k=1}^{n} \frac{(k^2+nk)}{n^3} = \dfrac{1}{n^3} \times \dfrac{n(n+1)(2n+1)}{6} + \dfrac{1}{n^2} \times \dfrac{n(n+1)}{2}\quad \mathop{\longrightarrow}\limits_{n \to \infty}\quad \dfrac{2}{6}+ \dfrac{1}{2} = \dfrac{5}{6}$$
• Please explain what you have done here....instead of $k$ you have written some terms in $n$, how and why... Nov 20 at 16:29
• These are the basic formulas $$\sum_{k=1}^n k = \dfrac{n(n+1)}{2} \quad \text{and} \quad \sum_{k=1}^n k^2 = \dfrac{n(n+1)(2n+1)}{6}$$ If you don't know these formulas, I suggest that you try to prove them (by induction for example). Nov 20 at 16:31
$$S_n=\frac{1}{n}\sum_{k=1}^n\frac{k^2}{n^2}+\frac{1}{n}\sum_{k=1}^n\frac{k}{n}.$$