Question

Σ (1 + sin n)/10^n from 1 to infinity, does it converge or diverge? Σ (1 + sin n)/10^n from 1 to infinity, does it converge or diverge? @Mathematics

Answer 1

converge

Answer 2

try to break this up into components.
so the numerator is (1+sin(n)). since sin(n) is always between -1 and 1, then you know that the numerator must always be between 0 and 2.
the denominator is 10^n which increases rapidly as n increases.
so - do you have an idea now of what the sum would do?

Answer 3

it'd approach zero because the limit as it approaches infinity is 0?

Answer 4

correct!
each term of this sum will get smaller and smaller - so you would be adding less and less each time. therefore the sum should converge.

Answer 5

but i think i am supposed to use the direct comparison test to solve it

Answer 6

well you have done that in an "indirect" way
you could split up the sum into two sums:
1. Sum[1/10^n]
2. Sum[sin(n)/10^n]
the 1st sum definitely converges.
therefore the 2nd one also converges as its elements are less than or equal to the elements of the 1st sum

Answer 7

to prove that the 1st sum converges:
\[\sum_{1}^{\infty}\frac{1}{10^{n}}=\frac{1}{10}+\frac{1}{10^{2}}+...\]\[=\frac{1}{10}(\frac{1}{10^{0}})+\frac{1}{10}(\frac{1}{10})+\frac{1}{10}(\frac{1}{10^{2}})+...\]this is a geometric series with a=1/10, r=1/10, so the sum to "n" terms can be written as:
\[\frac{a(1-r^{n+1})}{1-r}=\frac{10^{n+1}-1}{9*10^{n+1}}=\frac{1}{9}-\frac{1}{9*10^{n+1}}\]as "n" tends to infinity, this sum tends to 1/9.
therefore this sum converges.