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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!

Thanks in advance!!

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  • $\begingroup$ 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. $\endgroup$
    – Stef
    Nov 21 at 2:13

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$$\sum_{k=1}^{n} \frac{(k^2+nk)}{n^3} = \dfrac{1}{n^3} \times \dfrac{n(n+1)(2n+1)}{6} + \dfrac{1}{n^2} \times \dfrac{n(n+1)}{2}\quad \mathop{\longrightarrow}\limits_{n \to \infty}\quad \dfrac{2}{6}+ \dfrac{1}{2} = \dfrac{5}{6}$$

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  • $\begingroup$ Please explain what you have done here....instead of $k$ you have written some terms in $n$, how and why... $\endgroup$
    – math student
    Nov 20 at 16:29
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    $\begingroup$ 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). $\endgroup$
    – TheSilverDoe
    Nov 20 at 16:31
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Hint

$$S_n=\frac{1}{n}\sum_{k=1}^n\frac{k^2}{n^2}+\frac{1}{n}\sum_{k=1}^n\frac{k}{n}.$$

Then use Riemann sum.

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