# Prove that $\left \{ \frac{n}{n+1}\sin\frac{n\pi}{2} \right \}$ is divergent.

I want to prove whether the sequence $\{a_n\} = \left \{ \dfrac{n}{n+1}\sin\dfrac{n\pi}{2} \right \}$ (defined for all positive integers $n$) is divergent or convergent.

I suspect that it diverges, because the $\sin\frac{n\pi}{2}$ factor oscillates between -1, 0 and 1.

Here is my attempt to prove that it is divergent:

Suppose that the sequence is convergent. This should lead to a contradiction. If it is convergent, then, for every $\epsilon>0$, there is an $N>0$ such that if $n>N$, then $\left|\dfrac{n}{n+1}\sin\dfrac{n\pi}{2} -L\right|<\epsilon$.

In particular, for $\epsilon = \frac{1}{2}$:

$$-\frac{1}{2}<\frac{n}{n+1}\sin\frac{n\pi}{2} -L<\frac{1}{2} \text{ for every integer }n>N$$

Notice that the sequence of values of $\sin\dfrac{n\pi}{2}$ is $1, 0, -1, 0, 1, 0, -1, \cdots$ for $n=1, 2, 3, 4, 5, 6, 7, \cdots$. If $n = 1, 5, 9 \cdots$, $\sin\frac{n\pi}{2}$ is $1$, so, if we choose a particular $n$ from this list of values, the value of $a_n$ is $\dfrac{n}{n+1}$. In this case, $\sin\dfrac{(n+2)\pi}{2}$ will be $-1$, therefore $a_{n+2} = -\dfrac{n+2}{n+3}$. So, we have that:

$$-\dfrac{1}{2}<\dfrac{n}{n+1}-L<\dfrac{1}{2} \text{ and } -\dfrac{1}{2}<-\dfrac{n+2}{n+3}-L<\dfrac{1}{2}$$

Rearranging the terms:

$$\dfrac{1}{2}>-\dfrac{n}{n+1}+L>-\dfrac{1}{2} \text{ and } \dfrac{1}{2}>\dfrac{n+2}{n+3}+L>-\dfrac{1}{2}$$

$$\dfrac{n}{n+1}+\dfrac{1}{2}>L>\dfrac{n}{n+1}-\dfrac{1}{2} \text{ and } \dfrac{1}{2}-\dfrac{n+2}{n+3}>L>-\dfrac{n+2}{n+3}-\dfrac{1}{2}$$

The first inequality says that $L$ is positive, but the second inequality says that $L$ is negative; if this is correct so far, then I found a contradiction, proving that the sequence is divergent.

My question is: is this correct or is there some inconsistency?

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This seems overly complicated. You should write out the first $9$ or so terms, and it'll be clear that there are 3 important subsequences here. Investigate the limit of each subsequence individually. – Ragib Zaman Jul 21 '12 at 0:37
It looks fine...though rather messy. – DonAntonio Jul 21 '12 at 0:38
Follow up on Ragib's suggestion. Note that the first factor $\frac{n}{n+1}$ converges to $1$. The terms $\sin(\frac{n\pi}{2})$ with $n$ congruent to $1$ mod $4$ are all $1$, while they are $-1$ if $n$ is congruent to $3$ mod $4$, and $0$ if $n$ is even. – Chris Leary Jul 21 '12 at 1:23

Claim for you to prove: if a sequence converges then any subsequence converges and to the very same limit as the whole sequence.

Well, what about choosing two very special subsequences here and proving they don't have the same limit? In fact, one of the "obvious" subsequences doesn't even have a limit...

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Here is another approach: If the sequence converges, then the difference between two successive terms $|a_{n+1}-a_n|$ must go to zero. So, if you can show that there are pairs of successive terms that do not go to zero, then the sequence diverges.

My suggestion is to consider pairs $a_{1+4k}, a_{2+4k}$, where $k$ is a positive integer. What is $|a_{1+4k} - a_{2+4k}|$, and does it converge to zero?

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Here is another approach:

show that there exists infinitely many terms within $\delta$ neighbourhood of 0,1 and -1.

in other words 0,1 and -1 are all limits so there is no single limit for this sequence.

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