# Thread: [SOLVED] simple series question

1. ## [SOLVED] simple series question

assume the following series converge: $\displaystyle \sum_{j=1}^\infty (d_{2j}+d_{2j+1})$ and $\displaystyle \sum_{j=1}^\infty (d_{2j+1}+d_{2j+2})$, where $\displaystyle d_j$ can have any sign, and the convergence is not necessarily absolute. In general this doesn't imply that $\displaystyle d_j\to 0$, but after a change of variables we may assume $\displaystyle d_j\to 0$. Now assuming this, is it true that $\displaystyle \sum_{j=1}^\infty d_j$ is convergent?

P.S. well, ok, I lied, it's not exactly a "simple question", but it shouldn't be too deep either....

2. Originally Posted by choovuck
assume the following series converge: $\displaystyle \sum_{j=1}^\infty (d_{2j}+d_{2j+1})$ and $\displaystyle \sum_{j=1}^\infty (d_{2j+1}+d_{2j+2})$, where $\displaystyle d_j$ can have any sign, and the convergence is not necessarily absolute. In general this doesn't imply that $\displaystyle d_j\to 0$, but after a change of variables we may assume $\displaystyle d_j\to 0$. Now assuming this, is it true that $\displaystyle \sum_{j=1}^\infty d_j$ is convergent?

P.S. well, ok, I lied, it's not exactly a "simple question", but it shouldn't be too deep either....
No. $\displaystyle d_j=(-1)^j$

3. Originally Posted by Drexel28
No. $\displaystyle d_j=(-1)^j$
no, I'm demanding $\displaystyle d_j\to 0$

(basically, if $\displaystyle d_{2j}\to c_1$ with some $\displaystyle c_1\ne0$, then $\displaystyle d_{2j+1}\to -c_1$, and we can make the change of variables $\displaystyle \widehat{d_{2j}}=d_{2j}-c_1$, $\displaystyle \widehat{d_{2j+1}}=d_{2j+1}+c_1$, and ask whether $\displaystyle \sum_{j=1}^\infty \widehat{d_{j}}$ is convergent)

4. I've not been formally trained in series, so I may be making a bad assumption, but if $\displaystyle d_j \to 0$, doesn't that imply $\displaystyle d_{2j} \to 0$?

5. Originally Posted by Turiski
I've not been formally trained in series, so I may be making a bad assumption, but if $\displaystyle d_j \to 0$, doesn't that imply $\displaystyle d_{2j} \to 0$?
yes, $\displaystyle d_j \to 0$ implies $\displaystyle d_{2j} \to 0$ and $\displaystyle d_{2j+1} \to 0$, but it's in no way helpful

6. If the sum:
$\displaystyle \sum_{j=1}^\infty (d_{2j}+d_{2j+1})$

is convergent then $\displaystyle \lim_{j \rightarrow \infty}{d_{2j}+d_{2j+1}}=0$

same for the second sum $\displaystyle \lim_{j \rightarrow \infty}{d_{2j+1}+d_{2j+2}}=0$

Now let k=2j

$\displaystyle \lim_{k \rightarrow \infty}{d_{k}+d_{k+1}}=0$ and $\displaystyle \lim_{k \rightarrow \infty}{d_{k+1}+d_{k+2}}=0$

Now use the ratio test for absolute convergence for series that need not have positive terms and need not be alternating to use this test.

7. This is how I approached this:

Let $\displaystyle S = \sum_{j=1}^{\infty} d_{2j} + d_{2j+1}, \ T = \sum_{j=1}^{\infty} d_{2j+1} + d_{2j+2}$ and $\displaystyle W_n = \sum_{j=1}^{n} d_j$.

Then the partial sums $\displaystyle S_n = \sum_{j=1}^{n} d_{2j} + d_{2j+1}$ and $\displaystyle T_n = \sum_{j=1}^{n} d_{2j+1} + d_{2j+2}$ converge.

Observe that $\displaystyle \forall n \geq 1, \ S_n = W_{2n+1} - d_1 \ \text{and } \forall n \geq 2, \ T_{n-1} = W_{2n} -d_1 -d_2$. Then from the convergence of $\displaystyle S_n , \ T_n$ we get that $\displaystyle W_{2n} \text{ and } W_{2n+1}$ converge (You can actually prove it with only having that $\displaystyle W_{2n}$ converges and $\displaystyle d_n \to 0$..)

Now you finish it

8. Originally Posted by Defunkt
This is how I approached this:

Let $\displaystyle S = \sum_{j=1}^{\infty} d_{2j} + d_{2j+1}, \ T = \sum_{j=1}^{\infty} d_{2j+1} + d_{2j+2}$ and $\displaystyle W_n = \sum_{j=1}^{n} d_j$.

Then the partial sums $\displaystyle S_n = \sum_{j=1}^{n} d_{2j} + d_{2j+1}$ and $\displaystyle T_n = \sum_{j=1}^{n} d_{2j+1} + d_{2j+2}$ converge.

Observe that $\displaystyle \forall n \geq 1, \ S_n = W_{2n+1} - d_1 \ \text{and } \forall n \geq 2, \ T_{n-1} = W_{2n} -d_1 -d_2$. Then from the convergence of $\displaystyle S_n , \ T_n$ we get that $\displaystyle W_{2n} \text{ and } W_{2n+1}$ converge (You can actually prove it with only having that $\displaystyle W_{2n}$ converges and $\displaystyle d_n \to 0$..)

Now you finish it
this works, thanks!