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**Showcase_22** This is a question that seems obvious except my method doesn't look quite right.

Here's what i've done so far:

$\displaystyle a_{2n} \rightarrow a$

$\displaystyle a_{2n+1} \rightarrow a$

N.T.S (need to show) $\displaystyle a_n$ converges.

$\displaystyle (a_{2n})^\infty_{n=1}$:

$\displaystyle a_2<a_4<a_6<........<a_{2n}<a$

$\displaystyle (a_{2n+1})^ \infty_{n=1}$

$\displaystyle a_3<a_5<a_7<.........<a_{2n+1}<a$

Adding these two expressions gives:

$\displaystyle a_2+a_3<a_4+a_5<a_6+a_7<.......<a_{2n}+a_{2n+1}<2a$

$\displaystyle \frac{a_2+a_3}{2}<\frac{a_4+a_5}{2}<\frac{a_6+a_7} {2}<.....<\frac{a_{2n}+a_{2n+1}}{2}<a$

From here i'm not sure where to proceed. I'm not even sure if this is the right way of doing it!

I have noticed that I can get all the terms except for $\displaystyle a_1$. However, I can't see a way of connecting the two subsequences apart from adding them together.

I was thinking of making $\displaystyle a_n=\frac{a_{2n}+a_{2n+1}}{2}$ and then I would have a nice increasing sequence that has a limit a. Otherwise i'm having some difficulty in seeing what to do next.

Could someone please give me a nudge in the right direction? I would be grateful if they could!

EDIT: I've just noticed that you can get a_1 if youmake n=0 for the second subsequence. I can't actually see how this helps unless I can somehow see the property that a_{2n+1} is less than a_n. Even then I would have trouble justifying the order a_1<a_2_a_3......etc