Finding the sum of an absolutely convergent series

Question

I have an absolutely convergent series $\sum_{n=1}^{\infty} a_n$.

It is given that $\sum_{n=1}^{\infty} a_{2n}=\dfrac{9}{8}$ and $\sum_{n=0}^{\infty} a_{2n+1}=\dfrac{-3}{8}$. I have to determine sum of series $\sum_{n=1}^{\infty} a_n$ with justification.

I am not comfortable with the justification part. Here is what I have tried:

As given series $\sum a_n$ is absolutely convergent, the two series formed by taking some terms from it must also be absolutely convergent. Thus the two series $\sum a_{2n}$ and $\sum a_{2n+1}$ must also be absolutely convergent. This means that sum of original series which is the sum of two absolutely convergent series must be equal to the sum of sums of two daughter series.

This is my justification. But I am not fully convinced with this.

Can anyone help me with a convincing justification for this? Any help will be greatly appreciated.

Answer 1

If $S_n=a_1+a_2+\cdots+a_n, U_n=a_1+a_3+\cdots+a_{2n+1}$ and $V_n=a_2+a_4+\cdots+a_{2n}$ then $S_{2n}=U_{n-1}+V_n$ and $S_{2n+1}=U_n+V_{n}$. This shows that $S_{2n}$ and $S_{2n+1}$ both converge to $\frac 9 8 -\frac 3 8=\frac 3 4$. This implies that $S_n \to \frac 3 4$ so $\sum a_n=\frac 3 4$. We don't even need absolute convergence for this!

Answer 2

Let $$b_n=\begin{cases}a_n,&(n=2k), \\0,&(n=2k+1)\end{cases}$$and $$c_n=\begin{cases}0,&(n=2k), \\a_n,&(n=2k+1).\end{cases}$$Note that $|b_n|\le|a_n|$ so $\sum b_n$ converges absolutely; similarly for $\sum c_n$. And $a_n=b_n+c_n$, so $\sum b_n+\sum c_n=\sum a_n$.

Edit: It's being said with some vehemence that absolute convergence has nothing to do with this. We don't need it here, but only because we're given that the two "subseries" converge.

False Fact: If $\sum a_n$ converges then $\sum a_n=\sum a_{2n}+\sum a_{2n+1}$.

True Fact: If $\sum a_n$ converges absolutely then $\sum a_n=\sum a_{2n}+\sum a_{2n+1}$.