Use $S_n-\frac{n}{n+1}T_n
For $n ≥ 1$, let $a_n$ be a real number greater than $0$. For $n ≥ 1$, we let $S_n=\sum_{i=1}^{n}a_i$ (so that $S_n$ is the n-th partial sum of $S_n=\sum_{i=1}^{n}a_i$) and we define $T_n = \frac{1}{n}\sum_{i=1}^{n}S_i$(the average of the first n partial sums).Suppose that the set $\{ja_j:j\ge1\}=\{a_1,2a_2,3a_3,...\}$ is bounded above by a real number M.
(a) Use induction to prove that for all $n ≥ 1$, we have $$S_n-\frac{n}{n+1}T_n<M$$
(b) Assume now, in addition to the above, that the sequence $\{T_n\}^\infty_{n=1}$ converges. Use this, together with the result from (a), to prove that the series$\sum_{n=1}^{\infty}a_n$ converges.

(a) Assume that the set $\{ja_j:j\ge1\}=\{a_1,2a_2,3a_3,...\}$ is bounded above by a real number $M$
Show $\forall n\in [1,\infty)\cap \mathbb{N}, S_n-\frac{n}{n+1}T_n<M$ by induction
Base case:
Let $n=1$, then by assumption we have:
$$S_n-\frac{n}{n+1}T_n=S_1-\frac{1}{1+1}T_1=\sum_{i=1}^{1}a_i-\frac{1}{2} \frac{1}{1}\sum_{i=1}^{1}S_i=a_1-\frac{1}{2}a_1=\frac{1}{2}a_1<M$$
Inductive steps:
Assume $S_k-\frac{k}{k+1}T_k<M$
Show $S_{k+1}-\frac{k+1}{k+2}T_{k+1}<M$
By assumption we have $$\sum_{i=1}^ka_i-\frac{k}{k+1}\frac{1}{k}\sum_{i=1}^k(\sum_{j=1}^ia_j)<M$$
$$a_1+\dots+a_k-\frac{1}{k+1}(a_1+(a_1+a_2)+...+(a_1+...+a_k))<M$$
$$a_1-\frac{ka_1}{k+1}+\dots+a_k-\frac{(1)a_k}{k+1}<M$$
$$a_1(1-\frac{k}{k+1})+\dots+a_k(1-\frac{1}{k+1})<M$$
$$a_1(\frac{1}{k+1})+\dots+a_k(\frac{k}{k+1})<M$$
Sicne $a_1>0$ and $a_1\le M$ implies $M>0$, then we have:
$$a_1(1)+\dots+a_k(k+1)<M(k+1)$$
Also $a_{k+1}(k+1)\le M$, that
$$a_1(1)+\dots+a_k(k)+a_{k+1}(k+1)<M(k+2)$$
$$a_1(\frac{1}{k+2})+\dots+a_{k+1}(\frac{k+1}{k+2})<M$$
$$a_1(1-\frac{k+1}{k+2})+\dots+a_{k+1}(1-\frac{1}{k+2})<M$$
$$a_1-\frac{(k+1)a_1}{k+2}+\dots+a_k-\frac{(1)a_{k+1}}{k+2}<M$$
$$a_1+\dots+a_{k+1}-\frac{1}{k+2}(a_1+(a_1+a_2)+...+(a_1+...+a_{k+1}))<M$$
$$\sum_{i=1}^{k+1}a_i-\frac{k+1}{k+2}\frac{1}{k+1}\sum_{i=1}^{k+1}(\sum_{j=1}^ia_j)<M$$
$$\Rightarrow S_{k+1}-\frac{k+1}{k+2}T_{k+1}<M$$
Therefore $\forall n\in [1,\infty)\cap \mathbb{N}, S_n-\frac{n}{n+1}T_n<M$ hold by induction.
(b) Also assume $\{T_n\}^\infty_{n=1}$ converges
Show $\sum_{n=1}^{\infty}a_n$ converges
By assumption we have:
$$1.\forall \varepsilon>0,\exists n_0\in \mathbb{N},s.t. \forall n\in\mathbb{N}, n\ge n_0\rightarrow L-\varepsilon<T_n<L+\varepsilon$$
First we prove $$\exists c \in \mathbb{R},\forall n\in\mathbb{N},\left|T_n\right|<c$$
Since $\varepsilon=1$ also hold for 1.
Then $\exists n_0\in\mathbb{N}s.t.\forall n\ge n_0,\left|T_n-L\right|<1$
By Triangle Inequality that
$$ \left| T_n \right|- \left| L \right| \le \left| T_n-L \right| <1$$
$$\Rightarrow\left|T_n\right|<\left|L\right|+1$$
Let $c=\max\{\left|T_1\right|,\dots,\left|T_{n_0}\right|,\left|L\right|+1\}$ we can conclude
$$\exists c \in \mathbb{R},\forall n\in\mathbb{N},\left|T_n\right|<c$$
$$\Rightarrow\left|\frac n {n+1} T_n\right| < c$$
By (a) we have
$$S_n-\frac{n}{n+1}T_n<M$$
$$\Rightarrow S_n < c+M$$
And we know that:

Let $\{a_k\}_{k=1}^\infty$ be a sequence consisting of only non-negative terms, and define $\{S_n\}_{n=1}^\infty$ to be the sequence of partial sums for the infinite series $\sum_{k=1}^\infty a_k$. If there exists an $M\in\mathbb{R}$ such that $S_n\le M$ for all $n\in\mathbb{N}$ then the series converges.

Since $a_n>0$ which $a_n$ is non-negative, therefore:
$$\sum_{n=1}^{\infty}a_n \text{ is convergent}$$

Updates
thx @trancelocation and @Kavi Rama Murthy 's answers for (a) and (b), @trancelocation 's direct calculation method is great, but (a) requires to prove it by indction, so we still need to show the following hold:(?)
$$\sum_{i=1}^ka_i-\frac{k}{k+1}\frac{1}{k}\sum_{i=1}^k(\sum_{j=1}^ia_j)<M$$
$$\Rightarrow \sum_{i=1}^{k+1}a_i-\frac{k+1}{k+2}\frac{1}{k+1}\sum_{i=1}^{k+1}(\sum_{j=1}^ia_j)<M$$
I write a proof of (a) use induction, but it's too...long, is there simpler way to show it with induction?

Concerning a) (assuming that $M>0$):
The inequality can be rewritten as follows: 
$$S_n-\frac{n}{n+1}T_n<M \Leftrightarrow \boxed{(n+1)S_n- nT_n < (n+1)M}$$
The boxed inequality can be easily proved by direct calculation using the fact


*

*$\boxed{nT_n} = \sum_{k=1}^n S_k = \sum_{k=1}^n \sum_{i=1}^k a_i = \boxed{\sum_{i=1}^n (n-i+1)a_i}$
To see this just note that the sum of the first $n$ partial sums consists of $n$ times $a_1$, $n-1$ times $a_2$ and so on till $1$ time $a_n$.
Now, you get immediately
$$(n+1)S_n - nT_n = \sum_{i=1}^n \left((n+1) - (n+1-i) \right)a_i$$ $$ = \sum_{i=1}^n ia_i \stackrel{ia_i \leq M}{\leq} nM < (n+1)M$$
