Evaluate limit of sums

Question

Evaluate $$\lim_{n \to \infty} \sum_{k=1}^n \frac{1}{n + \sqrt{k^2+n}}$$

I saw this as a Riemann Sum and tried to rewrite as $$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{1 + \sqrt{(\frac{k}{n})^2 + \frac{1}{n}}}$$

but I can't say this is equal to $$\int_0^1 \frac{1}{1 + \sqrt{x^2 + 1}}$$ because there would be a too big error.

The answer should be $\ln 2$.

Answer 1

Squeezing:

$$\frac1{1+\sqrt{\left(\frac kn\right)^2+\frac1n}}\stackrel{(0)}<\frac1{1+\sqrt{\left(\frac kn\right)^2+0}}=\frac1{1+\frac kn}$$

And it is obvious that:

$$\lim_{n\to\infty}\frac1n\sum_{k=1}^n\frac1{1+\frac kn}=\int_0^1\frac1{1+x}\ dx=\ln(2)$$

We then notice that:

$$\frac1{1+\sqrt{\left(\frac kn\right)^2+\frac1n}}\stackrel{(1)}>\frac1{1+\frac kn+\frac1{2k}}\stackrel{(2)}>\frac{1-\frac1{2k}}{1+\frac kn}\stackrel{(3)}>\frac1{1+\frac kn}-\frac1{2k}$$

And finally notice that:

$$\lim_{n\to\infty}\frac1n\sum_{k=1}^n\frac1{1+\frac kn}-\frac1{2k}\ge\ln(2)$$

which follows easily since

$$\frac1n\sum_{k=2}^n\frac1k<\frac1n\int_1^n\frac1x\ dx=\frac{\ln n}n$$

and this limit may be taken care of by L'Hospital's rule.

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$(0)$ is motivated by the inequality $\frac1a<\frac1b$ if $0<b<a$.

$(1)$ is motivated by taking the line tangent to $\sqrt{a+x}$ at $x=0$.

$(2)$ is motivated by taking the line tangent to $\frac1{1+a+x}$ at $x=0$.

$(3)$ is just some basic algebra.

Answer 2

A general approach for such "almost" Riemann sums is to evaluate as a double limit which, in this case, can be justified by uniform convergence of the inner limit:

$$\begin{align} \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{1 + \sqrt{(k/n)^2 + 1/n}} &= \lim_{n \to \infty} \lim_{m \to \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{1 + \sqrt{(k/n)^2 + 1/m}}\\ &= \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{1 + (k/n)} \\ &= \int_0^1 \frac{dx}{1 +x } \\ &= \log 2 \end{align}$$

The relevant theorem is if $x_{nm} \to y_n$ uniformly as $m \to \infty$ and $x_{nm}$ converges as $n \to \infty$, then the double limit exists and

$$\lim_{n \to \infty} x_{nn} = \lim_{n \to \infty} \lim_{m \to \infty}x_{nm} = \lim_{m \to \infty} \lim_{n \to \infty}x_{nm}$$

To show uniform convergence in this case, note that

$$\begin{align}\left|\frac{1}{n} \sum_{k=1}^n \frac{1}{1 + \sqrt{(k/n)^2 + 1/m}} - \frac{1}{n} \sum_{k=1}^n \frac{1}{1 + k/n} \right| &\leqslant \frac{1}{n} \sum_{k=1}^n \frac{|1 + k/n - (1 + \sqrt{(k/n)^2 + 1/m})|}{|1+ k/n||1 + \sqrt{(k/n)^2 + 1/m}|} \\ &\leqslant \frac{1}{n} \sum_{k=1}^n (\sqrt{(k/n)^2 + 1/m}-k/n) \\ &=\frac{1}{n} \sum_{k=1}^n \frac{1/m}{\sqrt{(k/n)^2 + 1/m} + k/n}\\ &\leqslant \frac{1}{n} \sum_{k=1}^n \frac{1/m}{1/\sqrt{m}} \\ &= \frac{1}{\sqrt{m}}\end{align},$$

showing that

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

uniformly for all $n$.

Another approach would be to set bounds and apply the squeeze theorem.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \lim_{n \to \infty}\sum_{k = 1}^{n}{1 \over n + \root{k^{2} + n}} & = \lim_{n \to \infty}\sum_{k = 1}^{n}\pars{% {1 \over n + \root{k^{2} + n}} - {1 \over k + n} + {1 \over k + n}} \\[5mm] & = \lim_{n \to \infty}\sum_{k = 0}^{n}\bracks{% {k - \root{k^{2} + n} \over \pars{n + \root{k^{2} + n}}\pars{k + n}} + {1 \over k + n}} \\[5mm] & = \lim_{n \to \infty}\sum_{k = 0}^{n}\bracks{% {1 \over k + n} - {n \over \pars{n + \root{k^{2} + n}}\pars{k + \root{k^{2} + n}}\pars{k + n}}} \end{align}

Note that

\begin{align} &0 < \verts{\sum_{k = 0}^{n} {n \over \pars{n + \root{k^{2} + n}}\pars{k + \root{k^{2} + n}}\pars{k + n}}} < n\,{n \over \pars{n + \root{1 + n}}\root{n}n} \\[5mm] \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,&\ {1 \over \root{n}} \,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\to}\,\,\,{\large 0} \end{align}

Then,

\begin{align} \lim_{n \to \infty}\sum_{k = 1}^{n}{1 \over n + \root{k^{2} + n}} & = \lim_{n \to \infty}\sum_{k = 0}^{n}{1 \over k + n} = \lim_{n \to \infty}\pars{{1 \over n}\sum_{k = 0}^{n}{1 \over 1 + k/n}} = \int_{0}^{1}{\dd x \over 1 + x} =\ \bbox[10px,#ffe,border:1px dotted navy]{\ds{\ln\pars{2}}} \end{align}