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Suppose that $\Sigma_{k=1}^\infty a_k$ converges. Prove that if $b_k\uparrow \infty$ and $\Sigma_{k = 1}^\infty a_kb_k$ converges, then

$b_m \Sigma_{k = m}^\infty a_k → 0$ as $m → \infty$.

Attemtp: Suppose $\Sigma_{k=1}^\infty a_k$ converges, and $\Sigma_{k = 1}^\infty a_kb_k$ also.Then we know by Abel's Formula that the sequences converge only iff its partial sums converge. If we let $\Sigma a_k = \Sigma \frac{a_kb_k}{b_k}$, then because $b_k$ is increasing we have $\frac{1}{b_k}$ is decreasing to zero. Then let $c_k = \Sigma_{ j = k}^{\infty} a_jb_j$. Then $\Sigma_{k = n}^m a_k = \Sigma \frac{a_kb_k}{b_k}= \Sigma_{k = n}^m \frac{c_k - c_{k+1}}{b_k} $$ = \Sigma_{k = n}^{m} \frac{a_kb_k}{b_k}$

Abel's Formula: Let $a_k,b_k$ be real sequences, and for each pair of integers $n \geq m \geq 1$ set $A_{n,m} = \Sigma_{k =m}^n a_k$. Then Abel's formula is given

$\Sigma_{k = n}^m a'_kb_k' = \Sigma_{k =m}^n a_k'b_n' - \Sigma_{k = m}^{n-1} \Sigma_{j = k}^m a_j'(b_{k+1}' -b_k')$ Then we can apply Abel's formula if we let $a_k' = a_kb_k$ and $b_k ' = \frac{1}{b_k}$.

Then $\Sigma_{k = n}^m a'_kb_k' = \Sigma_{k =m}^n a_k'b_n' - \Sigma_{k = m}^{n-1} \Sigma_{j = k}^m a_j'(b_{k+1}' -b_k') = \Sigma_{k =m}^na_kb_k\frac{1}{b_k} - \Sigma_{k = m}^{n-1} \Sigma_{j = k}^m a_jb_j(\frac{1}{b_{k+1}}\frac{1}{b_k})$

Can someone please help me ? I don't know how to simplify. I am suppose to Abel's formula. I would really appreciate it.

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Using summation by parts we have

$$\sum_{k=m}^Ma_k = \sum_{k=m}^M\frac1{b_k}a_kb_k= \frac1{b_M}S_M - \frac1{b_m}S_{m-1} + \sum_{k=m}^{M-1}\left(\frac1{b_k}-\frac1{b_{k+1}}\right)S_k,$$

where

$$S_k = \sum_{j=k}^{\infty} a_j.$$

Hence,

$$b_m\sum_{k=m}^Ma_k = \frac{b_m}{b_M}S_M - S_{m-1} + b_m\sum_{k=m}^{M-1}\left(\frac1{b_k}-\frac1{b_{k+1}}\right)S_k,$$

and

$$0 \leqslant \left|b_m\sum_{k=m}^Ma_k\right| \leqslant \left|\frac{b_m}{b_M}S_M\right| + |S_{m-1}| + |b_m|\sum_{k=m}^{M-1}\left(\frac1{b_k}-\frac1{b_{k+1}}\right)|S_k|.$$

If $m$ is sufficiently large then $b_m > 0$ and $|S_k| < \epsilon/2$ for $k \geqslant m$.

Whence,

$$0 \leqslant \left|b_m\sum_{k=m}^Ma_k\right| \leqslant \left|\frac{b_m}{b_M}S_M\right| + |S_{m-1}| + \frac{\epsilon}{2} b_m\left(\frac1{b_m}-\frac1{b_{M}}\right)\\ = \left|\frac{b_m}{b_M}S_M\right| + |S_{m-1}| + \frac{\epsilon}{2} \left(1-\frac{b_m}{b_{M}}\right).$$

Take the limit as $M \to \infty$. Then

$$\left|b_m\sum_{k=m}^{\infty}a_k\right| \leqslant |S_{m-1}| + \frac{\epsilon}{2} $$

If $m$ is sufficiently large, then $|S_{m-1}| < \epsilon/2$, and

$$\left|b_m\sum_{k=m}^{\infty}a_k\right| < \epsilon. $$

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  • $\begingroup$ You're welcome! $\endgroup$ – RRL Feb 26 '15 at 5:39

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