Compute the following limit, possibly using a Riemann Sum

Question

$$\lim _{n\to \infty }\sum _{k=1}^n\frac{1}{n+k+\frac{k}{n^2}}$$ I unsuccessfully tried to find two different Riemann Sums converging to the same value close to the given sum so I could use the Squeeze Theorem. Is there any other way to solve this?

Answer 1

It is easy to see that

$$ \begin{align} \left |\sum_{k=1}^n \frac{1}{n+k+k/n^2}-\sum_{k=1}^n \frac1{n+k}\right|&=\frac1{n^2}\sum_{k=1}^n \frac{k}{(n+k+k/n^2)(n+k)}\\\\ &\le \frac1{n^2}\sum_{k=1}^n \frac1k\tag 1 \end{align}$$

Then, using $\sum_{k=1}^n\frac1k =\gamma+\log(n) +O\left(\frac1n\right)$, we see that the limit of the left-hand side of $(1)$ as $n\to \infty$ is $0$.

Can you finish now?

Answer 2

Writing this as

$$\lim_{n \to \infty}S_{nn} = \lim_{n \to \infty}\frac{1}{n}\sum _{k=1}^n\frac{1}{1+\frac{k}{n}+\frac{k}{n}\frac{1}{n^2}} $$

a technique that often works is to evaluate the double limit

$$\lim_{m \to \infty} \lim_{n \to \infty} S_{mn} = \lim_{m \to \infty} \lim_{n \to \infty} \frac{1}{n}\sum _{k=1}^n\frac{1}{1+\frac{k}{n}+\frac{k}{n}\frac{1}{m^2}} = \lim_{m \to \infty}\int_0^1 \frac{dx}{1 +x + x/m^2} = \int_0^1 \frac{dx}{1 +x} $$

where the last step is justified by DCT.

We can justify $ \lim_{n \to \infty} S_{nn} = \lim_{m \to \infty} \lim_{n \to \infty} S_{mn} = \log 2$ by showing one of the iterated limits exhibits uniform convergence.

An example with more details is given here. The argument for justifying the use of the double limit in this case will be similar.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\lim _{n \to \infty }\sum _{k = 1}^{n}{1 \over n + k + k/n^{2}}} = \lim _{n \to \infty }\bracks{{n^{2} \over n^{2} + 1} \sum _{k = 1}^{n}{1 \over k + n^{3}/\pars{n^{2} + 1}}} \\[5mm] = &\ \lim _{n \to \infty }\braces{{n^{2} \over n^{2} + 1} \sum _{k = 1}^{\infty}\bracks{{1 \over k + n^{3}/\pars{n^{2} + 1}} - {1 \over k + n^{3}/\pars{n^{2} + 1} + n}}} \\[5mm] = &\ \lim _{n \to \infty }\braces{{n^{2} \over n^{2} + 1} \bracks{H_{\large n^{3}/\pars{n^{2} + 1} + n}\ -\ H_{\large n^{3}/\pars{n^{2} + 1}}}} \end{align} where $\ds{H_{z}}$ is a Harmonic Number.

In using the $\ds{H_{z}}$ asymptotic behavior as $\ds{\verts{z} \to \infty}$, it's straightforward found:

$$ \bbx{\lim _{n \to \infty }\sum _{k = 1}^{n}{1 \over n + k + k/n^{2}} = \ln\pars{2}} $$

Answer 4

I thought it might be instructive to present a method that relies only on the squeeze theorem. To that end, we proceed.

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First, it is trivial to see that

$$\sum_{k=1}^n \frac{1}{n+k+k/n^2}\le \sum_{k=1}^n \frac{1}{n+k}\tag 1$$

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Second, note that we have

$$\begin{align} \sum_{k=1}^n \frac{1}{n+k+k/n^2}&=\sum_{k=1}^n \frac{1}{(n+k)\left(1+\frac{k/n^2}{n+k}\right)}\\\\ &\ge \sum_{k=1}^n \frac{1}{(n+k)}\left(1-\frac{k/n^2}{n+k}\right)\\\\ &\ge \left(1-\frac1{n^2}\right)\sum_{k=1}^n \frac{1}{n+k}\tag2 \end{align}$$

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Putting $(1)$ and $(2)$ together reveals

$$\left(1-\frac1{n^2}\right)\sum_{k=1}^n \frac{1}{n+k}\le \sum_{k=1}^n \frac{1}{n+k+k/n^2}\le \sum_{k=1}^n \frac{1}{n+k}$$

whence application of the squeeze theorem yields the coveted limit

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

as expected!

Answer 5

The sum under limit is not a Riemann sum, but it differs from a Riemann sum by negligible amount.

To setup things let's note that by definition of Riemann integral we have $$\int_{0}^{1}f(x)\,dx=\lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^{n}f(t_k),\,\frac{k-1}{n}\leq t_k\leq \frac{k} {n} $$ It is easy to guess that the function $f$ involved here is given by $f(x) =1/(1+x)$ and let's choose $$t_1=0,t_k=\frac{k-1}{n}+\frac{k-1}{n^3},k=2,3\dots,n$$ The corresponding Riemann sum is $$S_n=\frac{1}{n}+\frac{1}{n}\sum_{k=2}^{n}\dfrac{1}{1+\dfrac{k-1}{n}+\dfrac{k-1}{n^3}}$$ which can be rewritten as $$S_n=\frac{1}{n}+\sum_{k=1}^{n-1}\frac{1}{n+k+(k/n^2)}$$ If the sum in question is denoted by $S'_n$ then we can see that $$S_n-S'_n=\frac{1}{n}-\frac{1}{n+n+(1/n^2)}$$ and clearly the above expression tends to $0$ as $n\to\infty $. The Riemann sum $S_n$ tends to $\int_{0}^{1}dx/(1+x)=\log 2$ and hence the given sum $S'_n$ also tends to $\log 2$.

The same technique has been used in this answer in a simpler manner.