# real analysis converging proof using Abel's formula.

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} $$= \Sigma_{k = n}^{m} \frac{a_kb_k}{b_k} Abel's Formula: Let a_k,b_k be real sequences, and for each pair of integers n \geq m \geq 1 set A_{n,m} = \Sigma_{k =m}^n a_k. Then Abel's formula is given \Sigma_{k = n}^m a'_kb_k' = \Sigma_{k =m}^n a_k'b_n' - \Sigma_{k = m}^{n-1} \Sigma_{j = k}^m a_j'(b_{k+1}' -b_k') Then we can apply Abel's formula if we let a_k' = a_kb_k and b_k ' = \frac{1}{b_k}. Then \Sigma_{k = n}^m a'_kb_k' = \Sigma_{k =m}^n a_k'b_n' - \Sigma_{k = m}^{n-1} \Sigma_{j = k}^m a_j'(b_{k+1}' -b_k') = \Sigma_{k =m}^na_kb_k\frac{1}{b_k} - \Sigma_{k = m}^{n-1} \Sigma_{j = k}^m a_jb_j(\frac{1}{b_{k+1}}\frac{1}{b_k}) Can someone please help me ? I don't know how to simplify. I am suppose to Abel's formula. I would really appreciate it. ## 1 Answer Using summation by parts we have$$\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.$\$

• You're welcome! – RRL Feb 26 '15 at 5:39