Question

Determine the convergence of E(n=0 to infinity), (1+sin(n))/10^n
Please help...I'm lost...Award At End!

Answer 1

Please note that 1+sin n is never smaller than 0 nor larger than 2.

Thus, we can compare this given series to the series \[\frac{ 2 }{ 10^{n} }\]

Answer 2

Determine whether or not the series \[\sum_{n=0}^{\inf}\frac{ 2 }{ 10^n }\] converges or not. if it does converge, then the given series also converges because the given series is always equal to or less than the comparison series.

Answer 3

Why is it never larger then 2?

Answer 4

I dont know how to determine it though...Thats what I am wondering..The steps :/

Answer 5

Take a look at the sine function. What is its range?

Now take that range and add 1 to both limits. What range do you get for 1+sin x?

Answer 6

You may find it helpful to make up your own list of various types of series and of various tests for the convergence of series.

Are you familiar with the integral test for convergence? Looking that up may be worth your while. Could you say, right off, whether or not \[\int\limits_{1}^{\inf}\frac{ 2 }{ 10^x }dx\] If this converges, what does that tell you about the related infinite series?

Answer 7

See: http://tutorial.math.lamar.edu/Classes/CalcII/IntegralTest.aspx

Answer 8

Conditions for convergence or divergence does not depend on the "first" number.

Sum of (1+sin n) / (10^n) with n = 0 to infinity. = (1+0)/(10^0) + Sum of (1+sin n) / (10^n) with n = 1 to infinity.

Sum of (1+sin n) / (10^n) with n = 0 to infinity. = 1 + Sum of (1+sin n) / (10^n) with n = 1 to infinity.

Answer 9

Correct?

Answer 10

@zerosniper123 note from the unit circle that -1<= sin(x) <= 1 for any value of x

now add 1 to everything

0<=sin(x) + 1<= 2

now divide by 10^n and notice that 10^n is always positive so the signs dont flip

0 <= (sin(x)+1)/10^n <= 2/10^n

now if you can show that 2/10^n converges you are done

hint: p-series

Answer 11

sorry not p-series, geometric series

Answer 12

not trying to jump on your toes, but the math processor is not showing and I think that makes it hard to follow, because we just see the code. maybe not...lol

Answer 13

Assuming that yours is a geometric series, does your common ratio pass the test for convergence of a geometric series?

Regarding your question: your series begins with n=0, so you may as well use n=0 as the lower limit of integration as you apply the Integral Test to 2/(10^n).