Let $\{a_n\}^\infty_{n=1}$ be a sequence of positive real number such that $\sum ^{\infty}_{n=1} a_n$ is convergent Let $\{a_n\}^\infty_{n=1}$ be a sequence of positive real number such that $\sum ^{\infty}_{n=1} a_n$  is convergent . Which of the following are convergent correct?
$$1. \sum ^{\infty}_{n=1} \frac{a_n}{1+a_n}$$
$$2. \sum ^{\infty}_{n=1} \frac{a_n^\frac{1}{4}}{n^\frac{4}{5}}$$
$$3. \sum ^{\infty}_{n=1} na_n\sin(\frac{1}{n})$$
1 is true beacuse $\lim _{n\to \infty} \frac{a_n}{1+a_n}$ is convergent
i dont how to processed for 2 and 3
 A: For 1)
From convergence we have that $a_n \to 0 \Rightarrow \exists n_0 \in \Bbb{N}$ such that $a_n<1 ,\forall n \geq n_0$
Thus $1+a_n>1 \Rightarrow \frac{a_n}{1+a_n} \leq a_n ,\forall n \geq n_0$
From this we can conclude that $\sum_n \frac{a_n}{1+a_n}$ converges.
For 2)
Use Holder inequality for $p=\frac{5}{4}$ and $q=4$ and you will conclude that the given sum converges.
But use the exponent $q=4$  for  $\frac{1}{n^{\frac{4}{5}}}$ and also use the fact that if $\sum_na_n$ converges then $\sum_na_n^p$  also converges for $p>1$ and $a_n>0 \forall n \in \Bbb{N}$
Thus 2) is correct.
For 3) 
Use the fact that $|\sin{x}| \leq |x| \Rightarrow na_n\sin{\frac{1}{n}} \leq a_n$ and then the comparison test gives the conclusion.
A: 3) is good: The formula $|\sin u|\leq|u|$, so $n\sin(1/n)\leq 1$ and hence $\displaystyle\sum na_{n}\sin(1/n)\leq\sum a_{n}<\infty$.
A: *

*Use limit comparison test $x_n=a_n$ and $y_n=\frac{a_n}{1+a_n}$. Hence convergent.

*Use Holder's inequality $p=4,q=\frac{4}{3}$ then $\sum_{n=0}^{\infty}\frac{a_n^{\frac{1}{4}}}{n^{\frac{4}{5}}}\le (\sum_{n=0}^{\infty}a_n)^{\frac{1}{4}}(\sum_{n=0}^{\infty} n^{\frac{16}{15}})^{\frac{3}{4}}$. hence converges.


3 Use limit comparison test $x_n=a_n$ and $y_n={na_n}{\sin(\frac{1}{n})}. $Hence convergent.
A: For 1 
simply note that $\frac{a_n}{1+a_n}\leq a_n$
