Prove divergence of the Series

[davidge]

Homework Statement

Given $b_n = 1 / n$ if $n$ odd and $b_n = 1 / n^2$ if $n$ even, show that the series $$\sum_{n=1}^{\infty} (-1)^n b_n$$ diverges.

Homework Equations

Did'nt find any for this problem

The Attempt at a Solution

I assumed that $\sum_{n=1}^{\infty} (-1)^n b_n = \sum_{n \ \text{even}} b_n - \sum_{n \ \text{odd}} b_n$.

If we make the substitution $n = 2k, k \in \mathbb{N}$ for $n$ even and $n = 2k + 1, k \in \mathbb{N} \cup 0$ for $n$ odd, then the original series becomes

$$\sum_{n=1}^{\infty} (-1)^n b_n = \sum_{k=1}^{\infty} \frac{1}{(2k)^2} - \sum_{k=0}^{\infty} \frac{1}{2k+1}$$
By using the integral test for divergence, we easily find that the second sum on the RHS diverges, what causes the divergence of the whole series.

The professor said it can't be done the way I did, because any rearrengement of the series changes the final result. But this doesn't seem true, since we are dealing with all the terms, How can the result be different?

BTW, this was an exam question and she gave me a zero, claiming that I was wrong on that.

 

[andrewkirk]

Rearranging the order of terms in a series can change the result if there are an infinite number of both positive and negative terms, which is the case here. A series that converges to the same limit irrespective of any rearrangement of terms is called unconditionally convergent.

The Riemann Series Theorem states that a conditionally convergent series can be rearranged to add to any limit at all, including $\pm\infty$. The link has some examples of how to do this.

 

[davidge]

Thanks.
So my solution is wrong?

 

[andrewkirk]

Unfortunately, yes. I think giving it a zero was a bit tough though!

A way to prove it without rearranging terms is to note that:

\begin{align*}
\sum_{n=1}^{\infty} (-1)^n b_n
&= \sum_{k=0}^{\infty}\left( (-1)^{2k+1} b_{2k+1} + (-1)^{2k} b_{2k}\right)\\
&= \sum_{k=0}^{\infty}\left( \frac1{2k} - \frac1{2k+1} \right)\\
&= \sum_{k=0}^{\infty}\left( \frac1{2k} - \frac1{(2k+1)^2} \right)\\
&= \sum_{k=0}^{\infty}\left( \frac{2k+1-1}{(2k+1)^2} \right)\\
&= \sum_{k=0}^{\infty}\left( \frac{2k}{(2k+1)^2} \right)\\
&\geq \sum_{k=1}^{\infty}\left( \frac{2k+1}{2(2k+1)^2} \right)
\end{align*}
and, since that series has all positive terms, they can be rearranged without changing convergence. A judicious rearrangement shows that the sum is greater than one quarter of $\sum{j=1}^\infty 1/j$, which is known to diverge.

 

[davidge]

This was very clarifying. Thank you very much.

I think giving it a zero was a bit tough though!

 

[Ray Vickson]

Homework Statement

Given $b_n = 1 / n$ if $n$ odd and $b_n = 1 / n^2$ if $n$ even, show that the series $$\sum_{n=1}^{\infty} (-1)^n b_n$$ diverges.

Homework Equations

Did'nt find any for this problem

The Attempt at a Solution

I assumed that $\sum_{n=1}^{\infty} (-1)^n b_n = \sum_{n \ \text{even}} b_n - \sum_{n \ \text{odd}} b_n$.

If we make the substitution $n = 2k, k \in \mathbb{N}$ for $n$ even and $n = 2k + 1, k \in \mathbb{N} \cup 0$ for $n$ odd, then the original series becomes

$$\sum_{n=1}^{\infty} (-1)^n b_n = \sum_{k=1}^{\infty} \frac{1}{(2k)^2} - \sum_{k=0}^{\infty} \frac{1}{2k+1}$$
By using the integral test for divergence, we easily find that the second sum on the RHS diverges, what causes the divergence of the whole series.

The professor said it can't be done the way I did, because any rearrengement of the series changes the final result. But this doesn't seem true, since we are dealing with all the terms, How can the result be different?

BTW, this was an exam question and she gave me a zero, claiming that I was wrong on that.

You can re-arrange finite sums with no problem, and doing that will work in this problem. Consider the finite sum $S_{2N+1}$ (the sum of the first $2N+1$ terms). We have
$$S_{2N+1} = [1/2^2 + 1/4^2+ \cdots + 1/(2N)^2] - [1 + 1/3 + 1/5 + \cdots + 1/(2N+1)].$$
The first (even) terms produce a convergent series, having a finite limit as $N \to \infty$, while the second (odd) terms produce a divergent series, diverging to $\infty$ as $N \to \infty$. Thus, $S_{2N+1}$ does not have a finite limit as $N \to \infty$. Actually, you would need to do a bit or work to show divergence of the odd terms, but it would not be too onerous.