Suppose $\sum\limits_{n=1}^\infty b_n$ diverges and $b_n\gt 0$, show that the series $\sum\limits_{n=1}^\infty \frac{b_n}{1+b_n}$ also diverges

Question

As the title says, given a series $b_n > 0$, where $\sum_{n=1}^\infty b_n$ is divergent:

Show that the series $$\sum_{n=1}^\infty \frac{b_n}{1+b_n}$$ is also divergent.

So I've defined the series as $\sum_{n=1}^\infty u_n v_n$ where $u_n = b_n$ and $v_n = \frac {1}{1+b_n}$.

I'm sure there was some sort of convergence test which stipulated that for a product of two series, if one diverges the entire thing does, however I might be mistaken. If so, how do I properly prove this?

similar question This may helps you to understand André Nicolas's answer. — Bumblebee, Mar 30 '15 at 05:42

score 8 · Accepted Answer · answered Mar 30 '15 at 05:23

8

If the sequence $(b_n)$ is unbounded, then $\frac{b_n}{1+b_n}$ is greater than $\frac{1}{2}$ for infinitely many $n$. So in particular $\frac{b_n}{1+b_n}$ does not have limit $0$, and therefore the series $\sum \frac{b_n}{1+b_n}$ diverges.

If on the other hand the sequence $(b_n)$ is bounded above by $B$, then $\frac{b_n}{1+b_n}\ge \frac{1}{B+1}b_n$ for all $n$, so by Comparison the series $\sum \frac{b_n}{1+b_n}$ diverges.

answered Mar 30 '15 at 05:23

André Nicolas

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I should've mentioned that the problem specified that $b_n$ was bounded. So the latter explanation makes sense. Thank you! – michaelhankin Mar 30 '15 at 05:33
2

You are welcome. The result does not require boundedness, so it is a good thing that you did not mention it. – André Nicolas Mar 30 '15 at 05:46

Suppose $\sum\limits_{n=1}^\infty b_n$ diverges and $b_n\gt 0$, show that the series $\sum\limits_{n=1}^\infty \frac{b_n}{1+b_n}$ also diverges

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