Question

Σ (1 + sin n)/10^n from 0 to infinity, determine the convergence...Can someone work out the answer? I'm completely lost on how to figure these out..AWARD AT END!

Answer 1

hmm

Answer 2

@mathmate Got any ideas?

Answer 3

You realize that by using the sandwich theorem, the sum is between 0/10^n and 2/10^n.
This way, you can get rid of the ugly sine term.

Answer 4

After that, you only have a GP to worry about, and the sum (of the limits) are not difficult to calculate.

Answer 5

Well wouldn't 0/10^n just be 0? And no I wouldn't know that...That's why I am asking for help

Answer 6

The sum of 0/10^n would evidently be zero, and that's the lower bound.

Answer 7

Okay the convergence is 20/9 then. It converges

Answer 8

Thanks for the help

Answer 9

It's the 2/10^n that you need to calculate as the sum of a geometric progression, with common ration 10.

Answer 10

*ratio

Answer 11

Do you think you could help me solve the integral of convergence of sqrt(1+x)...I'm studying these for my class this week and I can't figure them out.

Answer 12

I can try, if you post the question. Not sure if I can do it.

Answer 13

Well that is the question. Find the power series and interval of convergence of sqrt(1+x). I already used the Taylor Series to get the "power series" but I can't get the interval of convergence. I keep on getting lost in the steps. I need someone to show me them.

Answer 14

Use integral test:
[\int\limits_{1}^{\infty} \frac{ 1 }{10 ^{x}}+\frac{ \sin x }{10 ^{x} }dx\]
You will find that it tends to .0827487 (Wolfram Alpha, or you can integrate the terms by hand).
According to the integral test if the integral is convergent then
\[\sum_{1}^{\infty} \frac{ 1+\sin n }{ 10^{n} }\]
is also convergent.

Therefore \[\sum_{0}^{\infty}\frac{ 1+\sin x }{10^{x} }\]
is also convergent because when x=0, the summand isfinite.

Answer 15

Thank you @DominicNg :) That really does help. Wolfram Alpha is a great math tool! Wish they didn't charge for it. -_-

Answer 16

@dominicNg
It is better than you think, because the summand is not infinite when x=0. 10^0 turns out to be 1. :)

Answer 17

It is convergent for any n.

Answer 18

@zerosniper123
So did you expand sqrt(1+x), and what did you get.

Answer 19

The interval of convergence is the range of x for which the series will converge.

Answer 20

I don't know why I can't paste it...I used a math program to write it out...Im just gonna a pic of it.

Answer 21

@mathmate when I expanded it and found the power series using Taylors method I got that.

Answer 22

You need to show convergence for the maximum range of x.
Since it is an alternating series, convergence is easier to prove.

Answer 23

Okay, well could you do the steps?? Right now I am brain fried...I don't really get this stuff...I understand taylor series and power series and Maclaurin...But trying to take the integral.

Answer 24

The series you gave me is correctly expanded. This is the very first step.

Answer 25

Whats the next step?

Answer 26

Are you asked to find the interval of convergence using the integral test?

Answer 27

Its not asking me to use the integral test, just find the interval of convergence. @mathmate

Answer 28

Radius of convergence is 1.

Interval of convergence is (-1,1).

See enclosed.

Answer 29

@DominicNg thank you for the detailed solution.
I hope you wouldn't mind if @zerosniper123 would ask for explanations on the solution.

Answer 30

What would you like to know