Study the convergence of $\sum_{n=1}^{\infty}\Bigl( \sqrt[n]{1+\frac{1}{n}}-1\Bigr)$

Question

I need to study the convergence of

$$ \sum_{n=1}^{\infty}\biggl( \sqrt[n]{1+\frac{1}{n}}-1\biggr). $$

Any help appreciated!!

Thanks!

Answer 1

By the binomial theorem, $$1\le 1+\frac 1n\le (1+\frac 1 {n^2})^n=1+n\frac 1 {n^2} +\binom{n}{2} (\frac 1 {n^2})^2+\dots,$$ so, taking $n$th roots, $$1\le \sqrt[n]{1+\frac{1}{n}}\le 1 + \frac1 {n^2},$$ and the sum converges by the comparison test.

Answer 2

Hint We have $$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\cdots +y^{n-1}).$$ Let $x=\sqrt[n]{1+\frac{1}{n}}$ and let $y=1$. Multiply top and (missing) bottom of your expression by $x^{n-1}+\cdots+y^{n-1}$, and look. The top is now nice and simple, and there is a big bottom.

Answer 3

So you've only got clever algebraic arguments so far. Here is an average analytic approach for the sake of completeness.

I will use two ingredients.

1) By concavity of $\log$, we have $\log(1+x)\leq x$ for all $x>-1$.

2) By the mean value theorem, $e^x-1\leq ex$ for all $x\in [0,1]$.

Now $$ 0\leq \sqrt[n]{1+\frac{1}{n}}-1=e^{\frac{1}{n}\log\left(1+\frac{1}{n}\right)}-1\leq e^{\frac{1}{n^2}}-1\leq \frac{e}{n^2}. $$

So the series converges by comparison with the Riemann $p$-series $\sum_{n\geq 1}\frac{1}{n^2}$.

Answer 4

Show for $n$ sufficiently large,

$$\sqrt[n]{1 + \frac{1}{n}} < 1 + \frac{1}{n^2}$$

The convergence will follow from comparison test.

Answer 5

Yet another approach...

Using Bernoulli's Inequality, for $n\ge1$, we have $$ \begin{align} 1+\frac1n &=1+\frac{n}{n^2}\\ &\le\left(1+\frac1{n^2}\right)^n \end{align} $$ Taking $n^{\text{th}}$ roots gives $$ \left(1+\frac1n\right)^{1/n}\le1+\frac1{n^2} $$ Thus, $$ \begin{align} \sum_{n=1}^\infty\left[\left(1+\frac1n\right)^{1/n}-1\right] &\le\sum_{n=1}^\infty\frac1{n^2}\\ &=\frac{\pi^2}6 \end{align} $$ That is, the series converges by comparison and the $p$-Test.

Answer 6

So far we've seen an average analytic argument to follow up on the clever algebraic arguments. Here is a below average analytic argument.

Using l'Hôpital's rule,

$$ \begin{align*} \lim_{n\to\infty}\frac{\left(1+\dfrac{1}{n}\right)^{1/n}-1}{\dfrac{1}{n^2}} &=\lim_{x\to 0}\frac{(1+x)^x-1}{x^2}\\ &=\lim_{x\to 0}\frac{(1+x)^x\left(\dfrac{x}{1+x}+\log(1+x)\right)}{2x}\\ &=\lim_{x\to0}(1+x)^x\lim_{x\to 0}\left(\frac{1}{2(1+x)}+\frac{\log(1+x)}{2x}\right)\\ &=1. \end{align*}$$

Convergence follows from the limit comparison test.

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Alternatively, inspired by Hurkyl's comment about a common way to estimate roots near $1$ , let $f(x)=(1+x)^{1/n}$ with $n>1$ fixed. Then $f(0)=1$, $f'(0)=\dfrac{1}{n}$, and $f''(x)< 0$ for all $x\geq 0$, and it follows that $f(x)< 1+\dfrac{x}{n}$ for all $x>0$. In particular, $f\left(\dfrac{1}{n}\right) <1+\dfrac{1}{n^2}$.

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Here's a sketch of another way to answer this with the limit comparison test, without l'Hôpital.

Let $g(x)=(1+x)^x$. Then $g$ is analytic in neighborhood of $x=0$, with $g(0)=1$ and $g'(0)=0$. Hence $g(x)=1+a_2x^2+a_3x^3+\cdots$, and $h(x)=\dfrac{g(x)-1}{x^2}=a_2+a_3x+\cdots$ is also analytic in a neighborhood of $x=0$, with $\lim\limits_{x\to 0}h(x)=a_2$. Thus $\lim_{n\to\infty}\frac{\left(1+\frac{1}{n}\right)^{1/n}-1}{\frac{1}{n^2}}=a_2$, which implies that the series converges by limit comparison.