Finding an upper bound for $|2^{1/n}-1|$ when $n$ goes to infinity for comparison test of convergence

Question

This question is about finding a small enough upper bound, for example, an upper bound like 100, n, is meaningless. And I am not sure if this sentence is ambiguous to some readers, but "to judge whether the series would converge" has nothing to do with "I need the final answer about convergence immediately after knowing this upper bound". Even if my only goal is to konw convergence, I don't agree that if the upper bound does not work well meaning I should not try to know how to find such an upper bound.

I am trying to find an upper bound (not least upper bound, but small enough for me to judge whether the series would converge) for this expression alone $|2^{1/n}-1|$, not the sum.

I saw a solution saying:

“We can use mean value theorem to prove $|2^{1/n}-1|$ <= $2^{1/n}(\log 2) n^{-1}$”

I can understand that suppose f(n)=$|2^{1/n}-1|$=$2^{1/n}-1$, $f^{'}(n)$=$2^{1/n}(\log 2) n^{-2}$ . But I don't understand how does this lead to the upper bound.

Question number 2, if using mean value theorem does not work, then, in general, how do we find an upper bound for an expression(no need to be the least upper bound, but small enough for me to judge whether the series would converge).

Thanks!

Answer 1

If you just wanted an upper bound for your sequence, I repeat my first comment (which I deleted when reading your post more carefully and seeing you talked about comparison test and series): $$|2^{1/n}−1|=2^{1/n}−1≤2−1.$$ But this upper bound is not "small enough to judge whether the series would converge" (better said: to prove that it converges, since an upper bound will never tell you that a series diverges).

And there is no hope for such a "small enough upper bound" because the series actually diverges like the harmonic series, since $$2^{1/n}-1\sim\frac{\ln2}n.$$ This asymptotic equivalence is sufficient to prove the divergence of your series, no need for an explicit lower bound (otherly said: the limit comparison test is sufficient, no need for the direct comparison test).

It has the advantage to also prove that your sequence converges to $0,$ no need for an explicit upper bound. But for this simpler question, this asymptotic argument itself is overkill. Simply use the continuity at $0$ of $x\mapsto 2^x.$

Applying the mean value theorem instead is not reasonable, since it uses a stronger property of this function (differentiability). But if you still insist on applying it, don't define $f(t)=2^{1/t}-1$ (this would lead to apply the mvt between $n$ and... infinity!). Define $f(x)=2^x-1$ (or simply $2^x$) and apply the mvt between $1/n$ and $0$ (like @Riemann indicated in his comment above) or more generally between any $x>0$ and $0:$ you get $$0<2^x-1<x\,2^x\ln2.$$

Answer 2

The answer provided by Anne Bauval is the correct solution for this with respect to the series. If you're considering the series

$$ \sum_{k=1}^\infty\left(2^{\frac{1}{2}} - 1\right), $$ then you can show that

$$\frac{\ln2}n < 2^{1/n}-1, $$

and therefore,

$$ \sum_{k=1}^\infty\frac{\ln(2)}{k} < \sum_{k=1}^\infty\left(2^{\frac{1}{k}} - 1\right). $$

This is enough to conclude that the series diverges by the Comparison Test.

However, if you're trying only to show that the sequence

$$ \left\{2^{\frac{1}{k}} - 1\right\}_{k=1}^\infty $$

has an upper bound, then it is enough to show that the function is monotonically decreasing (the first derivative is always strictly less than zero when $x>0$) and is bounded above by $1$ when $k=1$.

I hope the problem leads you down this path to show you that just because the argument of the series converges to zero DOES NOT imply that the series converges; however, it is true that if the series converges, then the argument must converge to zero. The Harmonic Series is an excellent counterexample of the first statement.