# convergence of infinite series

• Apr 15th 2010, 01:22 PM
234578
convergence of infinite series
Determine whether the following series is convergent (hint: use the comparison test)

1-(1/2)+(2/3)-(1/3)+(2/4)-(1/4)+....
ie.
(2/2)-(1/2)+(2/3)-(1/3)+(2/4)-(1/3)+...

Am I allowed to turn this into
(1/2)+(1/2)+(1/4)+..... by summing each pair of terms in the series, or is that only allowed if I know the series converges absolutely?

If I can do that, then the series is infinitesum{ 1/(n+1) } which equals infinitesum{ 1/n } -1 and since infinitesum{1/n} diverges, so does infinitesum{ 1/n } -1 I would think.

But I didn't use comparison...
• Apr 15th 2010, 06:47 PM
tonio
Quote:

Originally Posted by 234578
Determine whether the following series is convergent (hint: use the comparison test)

1-(1/2)+(2/3)-(1/3)+(2/4)-(1/4)+....
ie.
(2/2)-(1/2)+(2/3)-(1/3)+(2/4)-(1/3)+...

Am I allowed to turn this into
(1/2)+(1/2)+(1/4)+..... by summing each pair of terms in the series, or is that only allowed if I know the series converges absolutely?

The second option (i.e., you can't arbitrarily group terms in a general series), but even then I can't see how woould you get what you wrote if you group in pairs...it should be, imo, $\displaystyle \frac{1}{2}+\frac{1}{3}+\frac{1}{6}+\ldots...$

Tonio

If I can do that, then the series is infinitesum{ 1/(n+1) } which equals infinitesum{ 1/n } -1 and since infinitesum{1/n} diverges, so does infinitesum{ 1/n } -1 I would think.

But I didn't use comparison...

.