Show convergence divergence. Given a convergent series

Let $\sum a_n , a_n \gt 0$ be a convergent series. Now which of following is true

1. Series $\sum \sqrt{a_n}$ is always convergent

2. Series $\sum (\frac{a_1+a_2+....+a_n}{n})$ is always divergent

My attempt

To show 1 is true i used root test, since $\lim (a_n)^{1/n}\lt 1$ so $\lim ((a_n)^{1/n})^{1/2}\lt 1$also so 1 is true

I showed 2 is also true. Let $x_n$ be nth term of series given in 2. Then $\lim \frac{x_n}{1/n}=\lim (a_1+ a_2+....+a_n)$ which is finite as it just partial sum of convergent series $a_n$ so by comparision test series in 2 behaves like series $\sum 1/n$ therefor diverges

My doubt :is above process correct?any other way will be appreciated

1. Suppose $a_n=1/n^2$, so $\sum a_n$ is convergent. Then $\sum\sqrt{a_n}=\sum1/n$ which is divergent. So 1. cannot be true since we have a counter example.
2. Let $s_n=\sum_1^n a_k$ and $s=\lim_{n \to \infty} s_n$, then since $a_n>0$ eventually $s_n>s/2$. So eventually $s_n/n\;>\; s/(2n)$, and so the series $\sum s_n/n$ diverges.
• @Foggy There is the possibility of $\sum a_n$ being convergent but $\lim \sup a_n^{1/n}=1$, when the series with terms $b_n=\sqrt{a_n}$ will inherit this property and so the convergence will not be determined by the root test. – Conrad Turner Feb 2 '15 at 5:52
• @Foggy Which second part, part 2.? It is just a clarification of your method. The partial sums of a convergent series of positive terms eventually exceed half the sum of the series. So in this case allows us to show that the terms series we are investigating eventually exceed a fixed multiple of the corresponding terms of the harmonic series. Then a comparison test ... By eventually I mean that there is a $N$ such that for all $n>N$ the property holds. – Conrad Turner Feb 2 '15 at 6:03