Question

Question #fe8cc

Answer 1

Suppose that:

`sum_(n=1)^oo a_n`

with `a_n > 0` is convergent. Then necessarily:

`lim_(n->oo) a_n = 0`

so we can find a number `N` such that:

`n > N => 0 < a_n < 1`

which implies:

`n > N => 0< a_n^2 < a_n`

Then by direct comparison also:

`sum_(n=1)^oo a_n^2`

is convergent, and also:

`sum_(n=1)^oo a_n^2+a_n`

is convergent, as it is the sum of two convergent series.

Conversely suppose that:

`sum_(n=1)^oo a_n^2+a_n`

is convergent. Clearly as `a_n^2 > 0` we have:

`0 < a_n < a_n+a_n^2`

and then by direct comparison also:

`sum_(n=1)^oo a_n`

must be convergent.