Prove that $\sum\limits_{n=0}^{\infty}{(e^{b_n}-1)}$ converges, given that $\sum\limits_{n=0}^{\infty}{b_n}$ converges absolutely.

Solution 1:

As $b_n\rightarrow 0$. There is some $N$ such that for every $n>N$, $|b_n|<1$. $$b_n\le e^{b_n}-1 =b_n+\frac{b^2_n}{2!}+\frac{b_n^3}{3!}+\cdots \le b_n+|b_n|\left(1+\frac{1}{2}+\frac{1}{2^2}+\cdots\right)\le b_n+2{|b_n|}$$ and hence $$|e^{b_n}-1|<\max\left\{|b_n|,|b_n+|2{b_n}|| \right\}<4|b_n|$$ Now use comparison test.

Solution 2:

A neat argument:

if $x\in [0,1]$, then by mean value theorem, $\displaystyle \left|\frac{e^x -e^0}{x} \right| \leq e^1$

That is to say, $|x|\leq 1 \implies |e^x-1|\leq e^1|x|$

Since $b_n\to 0$, there is some $N$ such that $n\geq N \implies |b_n|\leq 1$

Then, for fixed $M$, $\displaystyle \sum_{n=N}^M |e^{b_n}-1|\leq e^1\sum_{n=N}^M |b_n|$

Hence $\displaystyle \sum_{n\geq N} e^{b_n}-1$ is absolutely convergent and $\displaystyle \sum_{n\geq 1} e^{b_n}-1$ also is.

Solution 3:

The limit comparison test tell us that if $\sum |b_n|$ converges and $\lim |a_n|/|b_n| = L \neq 0$ then $\sum |a_n|$ converges.

Notice that since $\sum b_n$ converges then $b_n \to 0$. Now consider $$\frac{e^{b_n} - 1}{b_n}.$$

Since $b_n$ is going to zero, this means that the limit is the same as the derivative of $e^x$ at $x=0$. In other words: $$\frac{e^{b_n} - 1}{b_n} \to 1$$ and the series $\sum (e^{b_n} - 1)$ converges.

Solution 4:

Another one, using the equivalent test (aka the limit comparison test): since $\sum_n b_n$ is convergent, then $b_n\underset{n\to+\infty}\longrightarrow0$, hence $$\mathrm{e}^{b_n}-1\underset{n\to+\infty}\sim b_n,$$ hence $$\bigl\lvert\mathrm{e}^{b_n}-1\bigr\rvert\underset{n\to+\infty}\sim\lvert b_n\rvert\geq0,$$ hence by the equivalent test, the series $\sum_n\mathrm{e}^{b_n}-1$ converges absolutely.