Summation of infinite series

Question

If we know the series sum given below converges to a value $C$(constant) $$\sum_{n=0}^{\infty}a_n =C \tag 2$$ Can we generate following in terms of C. values of $a_n$ will tend to zero as n goes to infinity and sum too converges to C..

1. $\sum_{n=0}^{\infty}na_n $
2. $\sum_{n=0}^{\infty} \frac{a_n}{n!}$

Answer 1

no; let $a_n=2^{-(n+1)}$, $b_0=1, b_n=0$ otherwise $(n>0)$. $C=1$ in both cases, yet for $a_n$:
1. has value 1
2. has value $exp(1/2)$

while for $b_n$
1. has value 0
2. has value 1

EDIT: It seemed to me as if the only given was $C$, and not the general term of the series. There is no direct link between the value of C and the values of 1., 2. though. Simply permuting the elements of $a_n$ changes the values of both 1. and 2., so we see there are infinitely many values they can take for a given C. In fact, by constructing the series carefully, we can make 1. and 2. be anything given C (in infinitely many ways for each number we decide to make 1. and 2. equal to). Thus, in any case you wouldn't be relating the answer to C, but simply summing the resulting series and finding values 1. and 2. (unless you consider $(\sum_{n=0}^{\infty}na_n/C)*C$ a valid, meaningful relationship :D)
TL;DR C is irrelevant, all that matters is the general term.

However, Here is a trick for series 1.
Assuming everything is nice, write $f(x) = \sum_{n=0}^{\infty}na_nx^n$. We seek $f(1)$. Note that $f$ is the derivative of $F(x) =\sum_{n=0}^{\infty}a_nx^{(n+1)}$. If we can find $F$, then we can find $f$, and hence $f(1)$. This works with $a_n=2^{-(n+1)}$.

Answer 2

Counterexample for question #$1$:

• $ a_n= \begin{cases} 8 & \text{$n=0$}\\ 0 & \text{$n>0$}\\ \end{cases} $

• $\sum\limits_{n=0}^{\infty}a_n=8$

• $\sum\limits_{n=0}^{\infty}na_n=0$

So you cannot express $\sum\limits_{n=0}^{\infty}na_n$ in terms $C$ for every given $a_n$.