Ytukyg

$displaystylesum_{n=1}^{infty}na_{n}$ converges. Prove that:
$$
nsum_{k=n}^{infty}a_{k}to0
$$
Wish someone can provide me with some clue about this problem.

Nonnegativity is not required.

Suppose the series $sum k a_k$ converges and

$$lim_{n to infty} S_n = lim_{n to infty} sum_{k=1}^n k a_k = S$$

Summing by parts we get

$$n sum_{k=n}^m a_k = n sum_{k=n}^m ka_k frac{1}{k} = n left[frac{S_m}{m} - frac{S_{n-1}}{n} + sum_{k=n}^mS_kleft(frac{1}{k} - frac{1}{k+1} right)right]$$

The last sum on the RHS converges as $m to infty$ since the summand is $mathcal{O}(k^{-2})$. Taking the limit of both sides as $m to infty$ and noting that $S_m/m to 0$ we get

$$n sum_{k=n}^infty a_k = - S_{n-1} + nsum_{k=n}^infty S_kleft(frac{1}{k} - frac{1}{k+1} right)$$

For any $epsilon> 0$ and all sufficiently large $n$ we have for $k geqslant n$,

$$S- epsilon leqslant S_k leqslant S + epsilon,$$

and

$$-S_{n-1} +n (S- epsilon)sum_{k=n}^infty left(frac{1}{k} - frac{1}{k+1} right) leqslant nsum_{k=n}^infty a_k leqslant -S_{n-1} +n (S+ epsilon)sum_{k=n}^infty left(frac{1}{k} - frac{1}{k+1} right) $$

Noting the telescoping sum converging to $1/n$, it follows that

$$-S_{n-1} + S - epsilon leqslant nsum_{k=n}^infty a_k leqslant -S_{n-1} + S+ epsilon,$$

whence,

$$-epsilon leqslant liminf_{n to infty},, nsum_{k=n}^infty a_kleqslant limsup_{n to infty},,nsum_{k=n}^infty a_k leqslant epsilon$$

Since $epsilon$ can be arbitrarily close to $0$ we must have

$$lim_{n to infty} n sum_{k=n}^infty a_k = 0$$