# An example of series such that $0\le a_n\le b_n$ and $\sum a_n$ converges but $\sum b_n$ diverges

Looking to get some help with the following question.

I need to find nonnegative sequences $(a_n)$and $(b_n)$ such that $a_n\leq b_n$ for all $n$ and $\sum a_n$ converges but $\sum b_n$ diverges.

So the terms of $a_n$ must turn to $0$ but $b_n$ does not. But how do I find examples of this series?

• @Arrow thank you for the edit – mp12345 Jan 18 '17 at 23:13
• @SimpleArt sorry I had a typo there, just made a edit – mp12345 Jan 18 '17 at 23:15
• But it's not a question. – zhw. Jan 18 '17 at 23:16
• Why can't $b_{n}$ tend to zero? These terms can still constitute a divergent series. For example: $$a_{n} = {1 \over n^2}, \quad b_{n} = {1 \over n}.$$ – avs Jan 18 '17 at 23:17

Well, if $a_n=\frac1{2^n}$ and $b_n=n$, then

$$a_n\le b_n$$

$$\sum_{n=1}^\infty\frac1{2^n}=1<+\infty$$

$$\sum_{n=1}^\infty n=+\infty$$

But it is not necessary for $\require{cancel}b_n\cancel\to0$, for example, with $b_n=\frac1n$,

$$\sum_{n=1}^\infty\frac1n>\int_1^\infty\frac1x\ dx=\lim_{t\to\infty}\ln(t)=+\infty$$

But it is necessary for $a_n\to0$ by the term test.

• no I understand this now, thanks – mp12345 Jan 18 '17 at 23:20
• @mp12345 I updated, including avs's hint, in case you were wondering if it were necessary that $a_n$ and $b_n$ behave the way you predicted. – Simply Beautiful Art Jan 18 '17 at 23:22
• thanks again for your help – mp12345 Jan 18 '17 at 23:23
• @mp12345 No problem :-) – Simply Beautiful Art Jan 18 '17 at 23:23