Need help with summation/series.

**Diggidy** · April 13th 2009, 05:42 PM

(a) The series

= ?

b) The sequence

= ?

I have done the rest of the problems with out any problem.. but i cant seem to get this one right.

Please help,
Thank you

**apcalculus** · April 13th 2009, 06:22 PM

Originally Posted by Diggidy

(a) The series

= ?

b) The sequence

= ?

I have done the rest of the problems with out any problem.. but i cant seem to get this one right.

Please help,
Thank you

Need to think a bit about part a), but for now, here is some help with part b) if you set x = 2 / k (then note that x goes to zero as k goes to infinity.

The typical term of the sequence becomes A_k = (1 + x) ^(2/x) =
= [(1+x)^(1/x)]^2 whose limit, as x goes to zero, is e^2, where e is Euler's Constant. Some numerical investigation in Excel also verifies:

k A_k

1 3
2 4
3 4.62962963
4 5.0625
5 5.37824
6 5.618655693
7 5.8077951
10 6.191736422
15 6.536796047
100 7.244646118
1000 7.37431239
10000 7.387578632

Good luck!!

**Krizalid** · April 13th 2009, 06:41 PM

the series diverges as long as the limit of the sequence is finite.

**Soroban** · April 13th 2009, 06:51 PM

Hello, Diggidy!

$A_n \;=\;\left(1 + \frac{2}{n}\right)^n$

(b) The sequence $\{A_n\}\:=\:?$

apcalculus is correct . . .

Recall that: . $\lim_{z\to\infty}\left(1+\frac{1}{z}\right)^z \:=\:e$

We have: . $\left(1 + \frac{2}{n}\right)^n \;=\;\left(1 + \frac{1}{\frac{n}{2}}\right)^n \;=\;\bigg[\left(1 + \frac{1}{\frac{n}{2}}\right)^n\bigg]^{\frac{2}{2}} \;=$ . $\bigg[\left(1 + \frac{1}{\frac{n}{2}}\right)^{\frac{n}{2}}\bigg]^2$

If $n\to\infty$ , then $\tfrac{n}{2}\to\infty$

So we have: . $\lim_{\frac{n}{2}\to\infty}\left[ \left(1 + \frac{1}{\frac{n}{2}}\right)^{\frac{n}{2}}\right]^2 \;=$ . $\left[\lim_{\frac{n}{2}\to\infty}\left(1 + \frac{1}{\frac{n}{2}}\right)^{\frac{n}{2}}\right]^2 \;\;=\;\;e^2$

(a) The series: . $\sum^{\infty}_{n=1}A_n \;=\;?$

Since the $n^{th}$ term is does not go to zero, the series diverges.

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