There exists $N \in \Bbb N$ such that $a_n > a_{n+1}$ for all $n > N$. Suppose $a_n >0$ for all $n$ such that the series $\sum a_n$ converges.
Claim: There exists $N \in \Bbb N$ such that $a_n > a_{n+1}$ for all $n > N$.

Since the series $\sum a_n$ converges we know that $\lim a_n = 0$ and we have  $a_n >0$ for all $n$.
I am not able to go further.
 A: If $n$ is even let $$ a_n = \frac{1}{2^n}  $$
If $n$ is odd let $$ a_n = \frac{1}{3^n}  $$
A: One easy way to see why the claim is false: Take any monotonic strictly decreasing convergent series, for example
$$ a_n = \frac{1}{n^2}, \quad n \in \mathbb{N} $$
Then for all $n$ which is a power of $10$, you swap $a_n$ and $a_{n+1}$.
A: Why should it be true?
Consider $a_n = \frac 1{2^n}$ so $\sum_{k=1}^{\infty} a_k = 1$.
That's fine.  But now let $b_{2n-1} + b_{2n} = a_n = \frac 1{2^n}$.
We have $1 = \sum_{k=1}^\infty a_k = \sum_{k=1}^\infty (b_{2n-1} + b_{2n}) = \sum_{m=1;m\ odd}^\infty b_m + \sum_{m=2;m\ even}^{b_m} = \sum_{m=1}^{\infty} b_m$
so $\sum_{m=1}^\infty b_m =1$.
But can we conclude there is an $N$ so that $n > N\implies b_n >b_{n+1}$?  No.  We can have $b_{2n-1}$ but as small as we like (so long as it is over $0$)  and $b_{2n}$ be as large as we like (so long as $b_{2n}= \frac 1{2^n} - b_{2n-1}$).
For instance was can have $b_{2n-1} = \frac 1{100^n}$ and $b_{2n} = \frac 1{2^n} - \frac 1{100^n}$. Then we have $b_{2n-1} = \frac 1{100^n} < \frac {50^n-1}{100^n}= \frac 1{2^n} - \frac 1{100^n}= b_{2n}$.  Yet $\sum_{m=1}^{\infty} b_m =\sum_{j=1}^\infty (b_{2j-1} + b_{2j}) =(\frac 1{100^j} + \frac 1{2^j}-\frac 1{100^j}) = \sum_{m=j}\frac 1{2^j} = 1$.
The statement is false.
