Does the series $\sum_{n = 1}^{\infty}\left(2^{1/n} - 1\right)\,$ converge?

Question

I'm trying to determine if the following sum converges or diverges (this is question 38 in section 11.7 of Stewart's Early Transcendentals):

$$\sum_{n = 1}^{\infty}(2^{1/n} - 1)$$

I've considered all of the techniques I know (integration test, ratio test, etc.), but I upon inspection, none of them will solve this problem.

Any hints or help is appreciated.

Answer 1

According to the Mean Value Theorem $$ \mathrm{e}^h-1=h\,\mathrm{e}^{\vartheta h}, $$ for some $\vartheta\in(0,1)$.

In our case $$ 2^{1/n}-1=\mathrm{e}^{\ln 2/n}-1=\frac{\ln 2}{n}\mathrm{e}^{\vartheta \ln 2/n}>\frac{\ln 2}{n}, $$ since $\vartheta\in(0,1)$.

Hence the series $$ \sum_{n=1}^\infty \big(2^{1/n}-1\big), $$ diverges to infinity, by virtue of the Comparison Test, since so does series $$ \sum_{n=1}^\infty \frac{\log 2}{n}. $$

Answer 2

You may try to use Limit Comparison Test with $1/n$.

Edit:

Note that $2^{1/n}=e^{\frac{1}{n}\ln2}=1+\frac{1}{n}\ln2+\frac{(\frac{1}{n}\ln2)^2}{2!}+\frac{(\frac{1}{n}\ln2)^3}{3!}+\ldots$.

With this note, it is natural to think of the above mentioned test.

Answer 3

The function $x \mapsto 2^x - 1$ is strictly increasing and strictly convex; in particular, it lies above its tangent line at $0$ for all $x > 0$. Letting $f$ be the first order linear approximation from the tangent line, we have

$$f(x) = (2^0 - 1) + \ln 2 \left(x - 0\right) = x \ln 2$$

Therefore,

$$2^{1/n} - 1 > f\left(\frac 1 n \right) = \frac{\ln 2}{n}$$

Thus, by comparison to the harmonic series, the series is divergent.

Answer 4

It's worth noting that you can get a useful comparison without making use of calculus.

Let $2^{1/n}-1=a_n$. It's clear that $a_n\gt0$ for all $n$. Therefore, from

$$2=(1+a_n)^n=1+na_n+{n\choose2}a_n^2+\cdots+{n\choose n}a_n^n$$

we get

$$a_n={1\over n}\left(1-{n\choose2}a_n^2-\cdots-{n\choose n}a_n^n \right)\lt{1\over n}$$

But $a_n\lt1/n$ now implies

$$\begin{align} {n\choose2}a_n^2+{n\choose3}a_n^3+\cdots+{n\choose n}a_n^n &\lt{n(n-1)\over2!n^2}+{n(n-1)(n-2)\over3!n^3}+\cdots+{n(n-1)\cdots2\cdot1\over n!n^n}\\ &\le{1\over2!}+{1\over3!}+\cdots+{1\over n!}\\ &\lt{1\over2}+{1\over6}\left(1+{1\over2}+{1\over4}+{1\over8}+\cdots \right)\\ &={5\over6} \end{align}$$

(Note, we could have gotten $e-2$ instead of $5/6$, but I'm trying to avoid anything that smacks of calculus, and all we really need is a bound less than $1$.)

Putting this back into the expression for $a_n$ gives

$$a_n={1\over n}\left(1-{n\choose2}a_n^2-\cdots-{n\choose n}a_n^n \right)\gt{1\over n}\left(1-{5\over6} \right)={1\over6n}$$

and that's all we need.

Answer 5

You can approach this problem with a simple inequality: if $a>b>0$, then: $$ n(a-b)\,b^{n-1}<a^n-b^n < n(a-b)\,a^{n-1},\tag{1}$$ so, if we choose $a=2^{1/n}$ and $b=1$, from the $RHS$ we get: $$ 2^{1/n}-1 > \frac{1}{n\cdot 2^{\frac{n-1}{n}}}>\frac{1}{2n}\tag{2}$$ so the series diverges by comparison with the harmonic series.