Infinite series

[asdf1234]

Hi can anyone help me with these problems please:

1. Find the sum of the first five terms of ∑ n=1 to infinity

(-1)^(n-1) (n+1)/(2)^n

How large is the error in using this approximation of the sum?


i think i found the sum to be .53125 but i don't know if i'm doing it correctly..and i have no idea how to find the error in using the sum.

2. ∑ n=0 to infinity n^(1/2) / ((n^2)+3) tell whether this converges absolutely, conditionally, or diverges. show what test you used.

3. find the interval of convergence for this series. be sure to check the endpoints and tell where it converges conditionally

∑ n=1 to infinity (3^n)(x^n) / (n+1)^2


thank you so much in advance!!!!

 

[tkhunny]

1. Find the sum of the first five terms of ∑ n=1 to infinity
(-1)^(n-1) (n+1)/(2)^n
How large is the error in using this approximation of the sum?
i think i found the sum to be .53125

There is nothing magic about it. Just do it.

First, do you REALLY not know what "sum of five terms" means?

Once you see it, it will seem simple.

n = 1 -- (-1)^(1-1) (1+1)/(2)^1 = 2/2 = 32/32
n = 2 -- (-1)^(2-1) (2+1)/(2)^2 = -3/4 = -24/32
n = 3 -- (-1)^(3-1) (3+1)/(2)^3 = 4/8 = 16/32
n = 4 -- (-1)^(4-1) (4+1)/(2)^4 = -5/16 = -10/32
n = 5 -- (-1)^(5-1) (5+1)/(2)^5 = 6/32

Sum of first 5? (32-24+16-10+6)/32 = 20/32 = 5/8 = 0.625

The exciting feature with an "Alternating" series is the ability to put a handle on the error from the infinite sum. Since the terms are decreasing in absolute value AND changing signs, the error must be less than the first term not included. In this case, that is n = 6, or -7/64.

 

[soroban]

Hello, asdf1234!

Here's #3 . . .

3. find the interval of convergence for this series.
Be sure to check the endpoints and tell where it converges conditionally

. . \(\displaystyle \L\sum^{\infty}_{n=1}\frac{(3^n)(x^n)}{(n\,+\,1)^2\)

Ratio Test

\(\displaystyle \L\left|\frac{a_{n+1}}{a_n}\right|\:=\:\left|\frac{3^{n+1}x^{n+1}}{(n\,+\,2)^2}\,\cdot\,\frac{(n\,+\,1)^2}{3^nx^n}\right| \:=\:\left|3x\cdot\left(\frac{n\,+\,1}{n\,+\,2}\right)^2\right| \:=\:\left|3x\cdot\left(\frac{1\,+\,\frac{1}{n}}{1\,+\,\frac{2}{n}}\right)^2\right|\)

Take the limit:

\(\displaystyle \L R\:=\:\lim_{n\to\infty}\left|3x\cdot\left(\frac{1\,+\,\frac{1}{n}}{1\,+\,\frac{2}{n}\right)^2\right|\:=\:\left|3x\cdot\left(\frac{1\,+\,0}{1\,+\,0}\right)^2\right| \:=\:|3x|\)


For convergence, \(\displaystyle R\,<\,1\)
. . So we have: \(\displaystyle \:|3x|\,<\,1\;\;\Rightarrow\;\;-1\:<\:3x\:<\:1\;\;\Rightarrow\;\;-\frac{1}{3}\:<\:x\:<\:\frac{1}{3}\)

. . Hence, the series converges absolutely on \(\displaystyle \;\left(-\frac{1}{3},\,\frac{1}{3}\right)\)


Check endpoints . . .

At \(\displaystyle x\,=\,\pm\frac{1}{3}\), we have: \(\displaystyle \L\:\sum^{\infty}_{n=1}\frac{3^n(\pm\frac{1}{3})^n}{(n\,+\,1)^2} \:=\:\sum^{\infty}_{n=1}\frac{(\pm1)^n}{(n\,+\,1)^2}\;\;\) . . . both series converge.


Therefore, the series converges absolutely on \(\displaystyle \left[-\frac{1}{3},\,\frac{1}{3}\right]\)

 

[asdf1234]

There is nothing magic about it. Just do it....

thanks for your help! i got the sum right i just messed up the last fraction when i did the problem. the whole "error" concept is still confusing to me..so what would the answer to the question be? is it just the next (6th) term then -7/64?

"How large is the error in using this approximation of the sum?"

Hello, asdf1234!

Here's #3 . . .

thanks so much soroban!!! i guess i did the problem right then! i wasn't sure if both of the endpoints were supposed to converge. i got the same answer as you but my teacher told me that most of the time only one would converge conditionally and the other would diverge so i was skeptical of my answer. thanks again you guys are awesome!

 

[asdf1234]

nvm i got the 2nd problem! thanks to everyone who helped