Prove if $\sum ^{\infty }_{n=1}\dfrac{a_{n}}{a_{n}+1}$ converges then $\sum ^{\infty }_{n=1}a_{n}$ converges

Question

I am trying to prove the following:

Let $\sum ^{\infty }_{n=1}a_{n}$ be a positive series ($a_{n}\ge0$). Prove that if $\sum ^{\infty }_{n=1}\dfrac{a_{n}}{a_{n}+1}$ converges then $\sum ^{\infty }_{n=1}a_{n}$ converges.

Thought of trying to prove that $\lim _{n\rightarrow \infty }a_{n}=0$ and then use: $\lim _{n\rightarrow \infty }\dfrac{a_{n}}{\left(\dfrac{a_{n}}{a_{n}+1}\right)}=\lim _{n\rightarrow \infty }(a_{n}+1)=1$ but I get stuck trying to prove that $\lim _{n\rightarrow \infty }a_{n}=0$ as well.

To be honest I am a bit lost. Any hints would be appreciated. Thanks!

Answer 1

Note $a_n$ is bounded, otherwise the series diverges. So $a_n<M$ for all n.

$$0\le\frac{a_n}{1+M}\le \frac{a_n}{1+a_n}$$

Since series is convergent, $\Rightarrow \lim_{n\to \infty}\frac{a_n}{1+a_n}=0$

By Squeeze theorem, $$\lim_{n\to\infty}\frac{a_n}{1+M}=0\Rightarrow\lim_{n\to\infty} a_n=0$$

-------(regarding $a_n$ is bounded)-------

Because $\lim_{n\to\infty} \frac{a_n}{1+a_n}=0$, so take $\epsilon=0.1$, there exists some $N>0$ such that for all $n>N$, $$|\frac{a_n}{1+a_n}−0|<0.1\Rightarrow a_n<\frac{1}9$$ So it is bounded.

Answer 2

Since $\sum_{n=1}^{\infty} \frac{a_n}{a_n +1}$ is convergent, $lim_{n\rightarrow\infty}\frac{a_n}{a_n +1}=0.$ So, $lim_{n\rightarrow\infty}\frac{a_n+1-1}{a_n +1}=0.$ From here we get $lim_{n\rightarrow\infty}\frac{1}{a_n +1}=1$ and $lim_{n\rightarrow\infty}(a_n +1)=1.$ Thus $lim_{n\rightarrow\infty}a_n=0.$ By comparison test, as you did, we conclude that $\sum_{n=1}^{\infty}a_n$ is convergent.

Answer 3

Let us define $b_{n}=\dfrac{a_{n}}{a_{n}+1}$. Clearly $b_{n}\to 0$ , otherwise the series wouldn't converge. Solve for $a_{n}$ and get

$a_{n}=\dfrac{b_{n}}{1-b_{n}}$ which immediately implies $a_{n}\to 0$.

Therefore, for $\epsilon$=$\dfrac{1}{10}$ there is some $n_{0}$ such that for $n\geq\,n_{0}$, we have $a_{n}<\dfrac{1}{10}$.

Now write $a_{n}=\dfrac{a_{n}}{a_{n}+1}(a_{n}+1)$. For $n\geq\,n_{0}$ we have $(1+a_{n})\,\leq\,(1+\dfrac{1}{10})$, i.e.

$1+a_{n}\,\leq\,\dfrac{11}{10}$. Thus, we obtain a finite sum plus a

sum which is $\leq$ $\sum_{n_{0}}^{+\infty}\dfrac{a_{n}}{a_{n}+1}(\dfrac{11}{10}) $ which clearly converges by assumption!! So the result is proved!

Answer 4

Write $$\frac {a_n} {a_n + 1} = \frac {a_n + 1} {a_n + 1} - \frac 1 {a_n + 1} = 1 - \frac 1 {a_n + 1}$$

Then $\dfrac {a_n} {a_n + 1} \to 0$ gives $1 - \dfrac 1 {a_n + 1} \to 0$. Rearranging gives the result.