[SOLVED] simple series question

**choovuck** · June 5th 2010, 09:03 PM

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

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

**Drexel28** · June 5th 2010, 09:28 PM

Originally Posted by choovuck

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

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

No. $d_j=(-1)^j$

**choovuck** · June 5th 2010, 09:42 PM

Originally Posted by Drexel28

No. $d_j=(-1)^j$

no, I'm demanding $d_j\to 0$

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

**Turiski** · June 5th 2010, 10:42 PM

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

**choovuck** · June 5th 2010, 11:31 PM

Originally Posted by Turiski

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

yes, $d_j \to 0$ implies $d_{2j} \to 0$ and $d_{2j+1} \to 0$ , but it's in no way helpful

**p0oint** · June 6th 2010, 02:44 AM

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

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

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

Now let k=2j

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

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.

**Defunkt** · June 6th 2010, 03:47 AM

This is how I approached this:

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

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

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

Now you finish it

**choovuck** · June 6th 2010, 08:08 PM

Originally Posted by Defunkt

This is how I approached this:

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

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

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

Now you finish it

this works, thanks!

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