Show a_n = 1 + 1/3 + 1/5 + ... +1/(2n-1) diverges

and

show b_n = 1 + 2x + 3x^2 + ... + nx^(n-1) converges to (1-x)^-2 for an x with absolute value less than 1.

Thanks!

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- Nov 3rd 2009, 11:25 AMSteve14Help with 2 sequences
Show a_n = 1 + 1/3 + 1/5 + ... +1/(2n-1) diverges

and

show b_n = 1 + 2x + 3x^2 + ... + nx^(n-1) converges to (1-x)^-2 for an x with absolute value less than 1.

Thanks! - Nov 3rd 2009, 02:23 PMJazz10
I was able to prove part A, any help with part B please?

- Nov 3rd 2009, 04:32 PMJazz10
Sorry, should have posted my answer to the first one. Still need help on second one.

a2 = 1 + 1/3

a5 = 1 + 1/3 + 1/5 + 1/7 + 1/9 > 1 + 1/3 + 1/9 + 1/9 + 1/9 = 1 + 1/3 + 1/3

a14 = 1 + 1/3 + (1/5 + 1/7 + 1/9) + (1/11 + 1/13 + 1/15 + 1/17 + 1/19 + 1/21 + 1/23 + 1/25 + 1/27) > 1 + 1/3 + 1/3 + (1/27 + 1/27 + 1/27 + 1/27 + 1/27 + 1/27 + 1/27 + 1/27 + 1/27) = 1 + 1/3 + 1/3 + 1/3

and you can continue to find a series of elements that are greater than 1/3. Since 1 + 1/3 + 1/3 + 1/3 + ... diverges, and a_n is greater than that sequence by the above, then a_n diverges. - Nov 3rd 2009, 05:46 PMredsoxfan325
Easier way to prove divergence of the first sum: Observe that

$\displaystyle \frac{1}{2}\sum_{k=1}^{\infty}\frac{1}{k}<\sum_{k= 1}^{\infty}\frac{1}{2k-1}$

so the comparison test says that the sum diverges.

For the second one, we know that the geometric series converges:

$\displaystyle 1+x+x^2+...+x^n+...=\frac{1}{1-x}$ for $\displaystyle |x|<1$

Differentiate both sides with respect to $\displaystyle x$ to achieve the desired result.