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

Question

It's a question from a test that I had, and I don't know how to prove this, so I am forwarding this to you.

$\sum \limits_{n=0}^{\infty }\:b_n$ is absolutely convergent series . How to prove that the series $\sum \limits_{n=0}^{\infty }\:(e^{b_n}\:-\:1)$ is also absolutely convergent?

We cannot assume here that $b_n\:\ge 0$ for every $n$.

Answer 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.

Answer 2

Hint: for sufficiently small $|x|$ and sufficiently large $C>0$ you have $$|e^x-1|\le |x|+C|x|^2.$$

Answer 3

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, $|x|\leq 1 \implies |e^x-1|\leq e^1|x|$

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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.

Answer 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.

Answer 5

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.

Answer 6

$$ \text{If $b$ is any real number then }e^b-1\ge b\tag 1 $$ and $$ \text{if $0<b\ll1$ then }e^b-1<2b. \tag 2 $$

From $(1)$ we get $0\le e^b-1\le b\le 2b$ if $1\gg b\ge0$ and from $(2)$ we get $|e^b-1|<|b|$ if $b<0$. Either way we have $$ |e^b-1|\le 2|b|\text{ if $b$ is close enough to $0$.} \tag 3 $$ And $b_n$ will be close enough to $0$ for all but finitely many $n$ because $b_n\to0$ as $n\to\infty$.

Thus $(3)$ tells us that $\sum_n (e^{b_n}-1)$ converges by a comparison test.

So how do we know that $(1)$ and $(2)$ are true?

The exponential function is concave upward since is second derivative is everywhere positive. Therefore it lies above all of its tangent lines. The line $b\mapsto b+1$ is tangent to $b\mapsto e^b$ at $b=0$, so lying above its tangent line means $e^b\ge b+1$. That gives us $(1)$.

Next notice that the line $b\mapsto 1+2b$ passes through the point $(0,1)$, and so does the curve $b\mapsto e^b$, but the line has a greater slope than the curve at that point. Therefore, immediately to the right of that point (i.e. $0<b\ll 1$) the line $b\mapsto 1+2b$ lies above the curve $b\mapsto e^b$. That gives us $(2)$.

(The mean value theorem is tacitly used here, since it is at the heart of the proofs of some basic facts about the relationship between derivatives and graphs.)