The limit of $S_n=\sum_{k=1}^{n} \frac{kn}{k+n^3}$

Question

Given $n>1$ and $S_n=\sum\limits_{k=1}^{n} \frac{kn}{k+n^3}$

Calculate $\lim\limits_{n \to \infty} S_n $

Well its obvious that $\frac{n^2}{n^3+n} \leq S_n\leq \frac{n^3}{n^3+1}$

$S_n$ converge to a limit $l$ such that $0\leq l \leq 1$

How can we determine the value of $l$ ?

Answer 1

This problem can be tackled by squeeze theorem plus riemann sum.

Note that $$S_n=\sum^n_{k=1}\frac{\frac{k}n}{\frac{k}{n^3}+1}\frac1n$$

We can obtain the following inequality easily

$$\sum^n_{k=1}\frac{\frac{k}n}{\frac{n}{n^3}+1}\frac1n\le S_n\le\sum^n_{k=1}\frac{k}n\frac1n$$

Taking the limit $n\to\infty$, the lower bound approaches $$\lim_{n\to\infty}\left(\frac1{\frac1{n^2}+1}\right)\int^1_0xdx=\frac12$$

The upper bound approaches $$\int^1_0xdx=\frac12$$ as well.

By squeeze theorem, $$\color{red}{\lim_{n\to\infty}S_n=\frac12}$$

(This agrees with your expectation that the limit is between $0$ and $1$.)

Answer 2

You can also get the result if you only estimate the denominator in order to get better bounds: $$\frac{1}{2} \stackrel{n\to\infty}{\longleftarrow}\frac{n(n+1)}{2(1+n^2)} = \frac{n}{n+n^3} \sum \limits_{k=1}^n k \leq S_n \leq \frac{n}{1+n^3} \sum \limits_{k=1}^n k = \frac{n^2(n+1)}{2(1+n^3)} \stackrel{n\to\infty}{\longrightarrow} \frac{1}{2}\, . $$

Answer 3

If you are familiar with harmonic numbers$$S_n=\sum_{k=1}^{n} \frac{kn}{k+n^3}=n^4 \left(H_{n^3}-H_{n^3+n}\right)+n^2$$ Now, using the asymptotics $$H_p=\gamma +\log \left({p}\right)+\frac{1}{2 p}-\frac{1}{12 p^2}+O\left(\frac{1}{p^4}\right)$$ and applying would give $$S_n=\frac{1}{2}+\frac{1}{2 n}-\frac{1}{3 n^2}-\frac{1}{2 n^3}+O\left(\frac{1}{n^4}\right)$$ which shows the limit and how it is approached.

Moreover, this gives a good approximation. For example, the exact value $$S_{10}=\frac{102209264496236445266975}{187134762733574325443631}\approx 0.546180$$ while the above expansion would give $$ \frac{3277}{6000}\approx 0.546167$$