Infinite sum converge to what value?

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pivoxa15
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Homework Statement


The infinite series (-1)^n(x/n) from n=1 converges. But what is the specific value of it?
 
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[tex]\ln(1+x) = \sum^{\infty}_{n=0} \frac{(-1)^n}{n+1} x^{n+1}[/tex]
[tex]\sum_{n=1}^{\infty} (-1)^n\frac{x}{n} = x\sum_{n=1}^{\infty} \frac{(-1)^n}{n}=x\log_e 2[/tex]
 
Gib Z said:
[tex]\ln(1+x) = \sum^{\infty}_{n=0} \frac{(-1)^n}{n+1} x^{n+1}[/tex]
[tex]\sum_{n=1}^{\infty} (-1)^n\frac{x}{n} = x\sum_{n=1}^{\infty} \frac{(-1)^n}{n}=x\log_e 2[/tex]

You have put x=1 so it should just be ln(2)?
 
But ln(2)>0 and [tex]\sum_{n=1}^{\infty} (-1)^n\frac{x}{n}<0[/tex] since the first term is negative and has the largest magnitude so will dominate the series. The series should equal -ln(2) so you may have made an error with your series manipulation.
 
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Yea sorry about that >.< I made a mistake with the starts of the series, some were n=1 and others n=0, and I didn't handle them well. But youve got the idea