3
$\begingroup$

Example of series such that $\sum a_n$ is divergent but $\sum \frac{ a_n}{1+ n a_n}$ is convergent.

I got one example from online that $a_n =\frac {1}{n^2} $for n is non square term and $a_n =\frac {1}{\sqrt n} $ for $n$ is square term . I know that $\sum a_n$ is divergent term but not convience with convergence of $ \sum \frac{ a_n}{1+ na_n}$. Any help will be appreciated.

$\endgroup$
5
$\begingroup$

The counterxample is correct. If $a_n =\frac {1}{n^2}$ when $n$ is non square term and $a_n =\frac {1}{\sqrt n} $ when n is square term, we have that $$\sum_{n=1}^{\infty} a_n=\sum_{\text{$n$ is not a square}} \frac{1}{n^2}+\sum_{\text{$n$ is a square}} \frac{1}{\sqrt{n}}\geq \sum_{k=1}^{\infty} \frac{1}{\sqrt{k^2}}=+\infty.$$ One the other hand $$\sum_{n=1}^{\infty} \frac{ a_n}{1+ n a_n}=\sum_{\text{$n$ is not a square}} \frac{1}{n^2+ n}+\sum_{\text{$n$ is a square}} \frac{1}{\sqrt{n}+ n}\\\leq \sum_{n=1}^{\infty} \frac{1}{n^2}+\sum_{k=1}^{\infty} \frac{1}{k+ k^2}\leq 2\sum_{n=1}^{\infty} \frac{1}{n^2}<+\infty$$

$\endgroup$
4
$\begingroup$

Delta-u and Robert Z have shown how to show convergence of $\sum{a_n\over1+na_n}$ for the example the OP was considering. Another example along the same lines is

$$a_n=\cases{1\quad\text{if }n=2^m\text{ for some } m\in\mathbb{N}\\ 0\quad\text{otherwise}}$$

In this case the sum $\sum a_n$ diverges because $a_n$ does not converge to $0$ (because it's infinitely often equal to $1$), while

$$\sum_n{a_n\over1+na_n}=\sum_m{1\over1+2^m}\lt\sum_m{1\over2^m}=2$$

The key idea in all the examples is to take a divergent series (like $1+{1\over2}+{1\over3}+\cdots$, in the OP's example, or, as here, $1+1+1+\cdots$) and space its terms more and more widely apart, with convergently small stuff (e.g., $1+{1\over4}+{1\over9}+\cdots$ or just plain $0+0+0+\cdots$) in between.

$\endgroup$
0
$\begingroup$

Hint

To show that $\sum \frac{ a_n}{1+ n \cdot a_n}$ is convergent notice that (as the terms are positive such decomposition is allowed): $$\sum \frac{ a_n}{1+ n \cdot a_n}=\sum_{n \text{ non-square}} \frac{\frac{1}{n^2}}{\frac{1}{n^2}+n\frac{1}{n^2}}+\sum_{m} \frac{\frac{1}{m}}{1+m^2\frac{1}{m}}=\sum_{n \text{ non-square}} \frac{1}{n^2+n}+\sum_{m} \frac{1}{m+m^2}$$

$\endgroup$
0
$\begingroup$

Slightly easier solution: Let $S=\{m^2:m\in \mathbb N\}.$ Define $a_n = 1$ if $n\in S,$ $a_n = 0$ otherwise. Then $\sum a_n$ diverges, since $a_n =1$ for infinitely many $n,$ hence $a_n\not \to 0.$ But

$$\sum_{n=1}^{\infty}\frac{a_n}{1+na_n} = \sum_{m=1}^{\infty} \frac{1}{1+m^2\cdot 1} \le \sum_{m=1}^{\infty} \frac{1}{m^2}<\infty.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.