Solving Nth Term of a Series Homework

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SUMMARY

The discussion focuses on solving the Nth term of a series defined by the nth partial sum Sn = 1/(n+1) + 1/(n+2) + ... + 1/(2n) = ∑(k=1 to n) 1/(n+k). The user seeks assistance in writing the associated series and understanding the general term. The associated series can be expressed as a_n = {1/2 for n=1, (1/(2n-1) + 1/(2n) for n≥2}. A suggestion is made to adjust the series by subtracting 1 to align the terms correctly.

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Homework Statement


Not really sure where tou go with this one.

Homework Equations


If the nth partial sum of a partial series is given by,

Sn= \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} = \sum_{k=1}^{n}\frac{1}{n+k}

a) write the associated series
b) test for convergence
c) if possible, determine its limit


The Attempt at a Solution



Here is what I have come up with:
s_1=1/2
s_2=1/2+1/3+1/4
s_3=1/2+1/3+1/4+1/5+1/6

a_n = \left\{ \begin{array}{c} \frac{1}{2} \text{ for }n=1 \\ \frac{1}{2n-1}+\frac{1}{2n} \text{ for }n\geq 2 \end{array} \right.<br />

I don't know what to do next. What do I do with the 1/2?

I am pretty sure I can handle b and c, I just need help with a.

Thanks in advance!
 
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You seem to have the general term. Really, you just now just say that associated series is sum of a_n, but if you want to put it "nicely", then putting in n=1 in the expression you have for n>2, its equal to 1.5, which is 1 too big. So you can use the same generating rule for all n, as long as you subtract 1 from the series.
 
"associated series"

Aimee79 said:
a) write the associated series

I am pretty sure I can handle b and c, I just need help with a.

Hi Aimee79! :smile:

The associated series just means the series {an} such that a1 + … + an = Sn.

Hint: subtract something. :wink:
 

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