1. ## sequences

Hi all,
I'm trying to solve the following 2 questions , can anyone help?
1. Let for k . Show that there exists a sequence 0 such that

2.Let .Show that there exists a sequence such that .

For the first part , I think for is suitable. Can anyone give a hint for the second part?

2. If $\displaystyle \displaystyle \sum_{k=1}^{\infty} |a_{k}|< \infty$, then the series converges absolutely and that means that $\displaystyle a_{k} \rightarrow 0$ and for k 'large enough' is $\displaystyle \displaystyle |a_{k}|< \frac{1}{k^{\alpha}}$ where $\displaystyle \alpha>1$. Now if we suppose that $\displaystyle \forall k, a_{k}\ne 0$ and set $\displaystyle b_{k}= \ln |a_{k}|$ it will be $\displaystyle |b_{k}| \rightarrow \infty$ and for k 'large enough' $\displaystyle \displaystyle |a_{k}\ b_{k}|< \frac{1}{k^{\beta}}$ where $\displaystyle \beta>1$ so that $\displaystyle \displaystyle \sum_{k=1}^{\infty} |a_{k}\ b_{k}|< \infty$ ...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

3. Originally Posted by chisigma
If $\displaystyle \displaystyle \sum_{k=1}^{\infty} |a_{k}|< \infty$, then the series converges absolutely and that means that $\displaystyle a_{k} \rightarrow 0$ and for k 'large enough' is $\displaystyle \displaystyle |a_{k}|< \frac{1}{k^{\alpha}}$ where $\displaystyle \alpha>1$.
Unfortunately that is not true. There exist convergent series $\displaystyle \sum a_n$ of positive numbers, such that the condition $\displaystyle a_n<\frac1{n^\alpha}$ does not hold for all 'large enough' n, for any $\displaystyle \alpha>1$. For example, if $\displaystyle a_n = \dfrac1{n(\ln n)^2}$, then $\displaystyle \sum a_n$ converges (by the integral test), but $\displaystyle n^\alpha a_n\to\infty$ for all $\displaystyle \alpha>1$.

Originally Posted by hermanni
Let .Show that there exists a sequence such that .
There is a standard trick to construct this example. Let $\displaystyle t_n$ denote the tail of the series $\displaystyle \sum|a_k|$, in other words $\displaystyle t_n = \sum_{k=n}^\infty|a_k|$. Then $\displaystyle t_n\to0$ as $\displaystyle n\to\infty$. Let $\displaystyle b_k = 1/\sqrt{t_k}$. Clearly $\displaystyle b_k\to\infty$ as $\displaystyle k\to\infty$.

For the clever proof that $\displaystyle \sum|a_kb_k|$ converges, see here.