# power series; 2

by arl146
Tags: power, series
 P: 343 because the 1/n adds more to the value than just the 1/n^2 on the bottom. but as n approaches infinity that value doesnt matter so both series are pretty much equal. is that right or am i thinking about this wrong?
 Emeritus Sci Advisor HW Helper Thanks PF Gold P: 11,378 The 1/n makes the denominator bigger in your series, so what does that imply about the terms of the series compared to 1/n2?
 P: 343 yea thats what i said didnt i? OHHHHHH i see what dumb thing i did. yea so the terms of my series is smaller than those of 1/n^2 sorry, i know i do overlook stuff easily like that. i just cant help it, my brain just does it no matter how hard i try so at x=5 i proved that it is convergent??? or is there some how more that im missing?
 P: 343 so.. did i actually prove it this time?
 Mentor P: 20,433 How about writing some mathematics instead of just describing things vaguely with words?
 P: 343 ok well we need $\frac{n}{n^3+1}$ < $\frac{1}{n^2}$ for the test and that can be proven by: $\frac{n}{n^3+1}$ = $\frac{n(1)}{n(n^2+1/n)}$ = $\frac{1}{n^2+1/n}$ so now we are looking at $\frac{1}{n^2+1/n}$ < $\frac{1}{n^2}$ for my series to be convergent since $\frac{1}{n^2}$ converges. looking at the denominators: n2+$\frac{1}{n}$ is > n2 thus making $\frac{n}{n^3+1}$ = $\frac{1}{n^2+1/n}$ < $\frac{1}{n^2}$ this true is that good enough?
 Mentor P: 20,433 Looks good!
 P: 343 ok great! so can i like write exactly what i have there for my homework?
Mentor
P: 20,433
What you wrote was fine, but a little more verbose and roundabout than necessary. This is how I would say it.
 Quote by arl146 ok well we need $\frac{n}{n^3+1}$ < $\frac{1}{n^2}$ for the test and that can be proven by: $\frac{n}{n^3+1}$ = $\frac{n(1)}{n(n^2+1/n)}$ = $\frac{1}{n^2+1/n}$
Now, since n2 + 1/n > n2, for all n >= 1,
then 1/(n2 + 1/n) < 1/n2, for all n >= 1.

Therefore, n/(n3 + 1) < 1/n2, for all n >= 1.
This shows that each term of the series in question is smaller than the corresponding term of the convergent p-series Ʃ1/n2.
 Quote by arl146 so now we are looking at $\frac{1}{n^2+1/n}$ < $\frac{1}{n^2}$ for my series to be convergent since $\frac{1}{n^2}$ converges. looking at the denominators: n2+$\frac{1}{n}$ is > n2 thus making $\frac{n}{n^3+1}$ = $\frac{1}{n^2+1/n}$ < $\frac{1}{n^2}$ this true is that good enough?
 P: 343 ok awesome. that is a lot shorter. and the endpoint check for x=3 is just the same concept so i think i got that. thanks a lot guys!
 P: 343 hmm i have another question (hopefully short) ... why do we check the endpoints for convergence? whats the purpose of doing that? do i always do that if the question asks for an interval of convergence and/or radius of convergence? does checking the endpoints change anything if its convergent or divergent?
 Mentor P: 20,433 The radius of convergence is the same whether the series converges at neither, either, or both endpoints. What does change is the interval of convergence, which could look like (a, b), (a, b], [a, b), or [a, b]. If the question asks for the interval of convergence, you need to check both endpoints.
 P: 343 ok yea i just figured that out right as i posted that.. so if they ask just to find the radius of convergence, since that answer comes before finding the interval of convergence, i wont have to check the endpoints since i dont even have to find the interval. correct?
 Mentor P: 20,433 Right

 Related Discussions Calculus & Beyond Homework 2 Calculus & Beyond Homework 5 Calculus & Beyond Homework 11 Calculus & Beyond Homework 11 Calculus & Beyond Homework 1