# If a series, $\sum^\infty_{n=1}a_n$, converges, does $\sum^{\infty}_{n=1}e^{a_n }-1$ converge as well?

It is my first instinct to assume this is true, but I'm not sure my work is sufficient to prove it. We begin with a convergent series $$\sum^\infty_{n=1} a_n$$ with positive, real terms. We want to determine if $$\sum^\infty_{n=1}(e^{a_n}-1)$$ is also convergent. My first idea was to first expand the exponent. This gives us $$\sum^{\infty}_{n=1} \sum^{\infty}_{k=0}\left(\frac{a_n^k}{k!}-1\right).$$ We can then "peel" off the zeroth term of the $$k$$ series giving us $$\frac{a_n^0}{0!} = 1$$. This cancels with the -1. Now, we are left with, $$\sum^{\infty}_{n=1} \sum^{\infty}_{k=1}\left(\frac{a_n^k}{k!}\right) .$$ Since we know $$a_n$$ converges, let's call its limit $$a$$. Now, $$\sum^{\infty}_{k=1}\left(\frac{a^k}{k!}\right) < \sum^{\infty}_{k=0}\left(\frac{a^k}{k!}\right) = e^a .$$ Thus we can conclude $$\sum^\infty_{n=1}e^{a_n}-1$$ is less than $$e^a$$ which is finite, thus it too must be finite.

Alternatively, one can use the limit comparison test. That is, for two series $$\sum s_n$$ and $$\sum a_n$$ with positive terms, $$\lim_{k\rightarrow\infty} \frac{s_k}{a_k} = \rho.$$ If $$\rho >0$$ then both series converge or both diverge. In this case we have, $$\lim_{n\rightarrow\infty}\frac{a_n} {e^{a_n}-1} = \frac{a}{e^a-1}$$ since our terms are positive and nonzero, clearly $$\rho > 0$$ thus both must converge.

• The answer to the question in the title is yes, because $e^x-1$ is a continuous function—this is a standard fact about all continuous functions. (Also $\{a_n\}$ is a sequence, not a series.) Jun 10, 2021 at 19:25
• oh, the OP really does mean series, not sequences? I was misled by the title Jun 10, 2021 at 19:55

If $$(a_n)$$ are assumed to be positive, your conjecture is true. If there is no other assumption, it's wrong.
Counterexample $$a_n= \frac{(-1)^n}{\sqrt{n}}$$.
Indeed, $$\sum a_n$$ is convergent, however $$\sum e^{a_n}-1 \stackrel{(*)}{\ge}\sum a_n+\frac{a_n^2}{2e} = +\infty$$ where in $$(*)$$ I have used the fact that $$e^x \ge 1+ x+ \frac{x^2}{2e}$$ for all $$x \ge -1$$