0
$\begingroup$

consider $ \displaystyle \sum_{n=1}^\infty (-1)^{n-1}a_n $ where $ (a_n) $ is a monotone decreasing sequence of nonnegative numbers with $ a_n \rightarrow 0 $ by the alternating series test, series of this form always converge.

Show that $ 0 \leq \displaystyle\sum_{n=1}^\infty (-1)^{n-1}a_n \leq a_1 $

There is a hint in the question:

if $ (S_N)$ denotes the sequence of partial sums, consider the subsequences $ (S_{2N}) $ and $ (S_{2N-1}) $ can you show that one is decreasing while the other is increasing?

I have proven the hint - however I am unable to proceed from here - Could someone please direct me to the correct direction?

$\endgroup$
0
$\begingroup$

We will show that $0\leq S_{n}\leq a_{1}$ for all $n\in\mathbb{N}$ (and hence $0\leq\lim S_{n}\leq a_{1}$).
We know that $0\leq a_{1}-a_{2}=S_{2}\leq S_{1}= a_{1}$ and you already showed that $S_{2n}$ is increasing while $S_{2n-1}$ is decreasing.
Now let $k\in\mathbb{N}$ and notice that $S_{2k}\leq S_{2k-1}$. Because $S_{2n}$ is increasing and $S_{2n-1}$ is decreasing we now know that $0\leq S_{2}\leq S_{2k}\leq S_{2k-1}\leq S_{1}=a_{1}$. To finish note that, for any $n\in\mathbb{N}$, $S_{n}=S_{2k}$ or $S_{n}=S_{2k-1}$ for some $k\in\mathbb{N}$.

$\endgroup$
0
$\begingroup$

there is two possibilities for n. Let n be odd. Then

$$S_{n}=a_1\underbrace{-a_2+a_3}_{\leq 0}\underbrace{-a_4+a_5}_{\leq 0}\cdot\cdot \cdot\underbrace{-a_{n-1}+a_{n}}_{\leq 0}\leq a_1$$.

On the other hand if n is even than $$S_{n}=a_1\underbrace{-a_2+a_3}_{\leq 0}\underbrace{-a_4+a_5}_{\leq 0}\cdot\cdot \cdot\underbrace{-a_{n-2}+a_{n-1}}_{\leq 0}\underbrace{-a_{n}}_{\leq 0}\leq a_1$$ So in any case $S_n\leq a_1$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.