infinite series

[squaremeplz]

1. The problem statement, all variables and given/known data

a) consider the infinite series (k=1) sum (inf) [(k+1)^(1/2) - (k)^(1/2)]
expand and simplify the nth partial sum. determine wether the oartial sums S_n converge as n-> inf

b) determine all the numbers x in R so that the infinite series

(k=0) sum (inf) [x^(k)/(k!)]

converges.

c) determine wheter the series

(k=1) sum (inf) [k/(k^3 + 1)] converges or diverges.




2. Relevant equations



3. The attempt at a solution

a) I wrote out the terms of the nth partial sums

S_1 = (2)^(1/2) - 1
S_2 = (2)^(1/2) - 1 + (3)^(1/2) - (2)^(1/2) = (3)^(1/2) - 1
S_3 = (2)^(1/2) - 1 + (3)^(1/2) - (2)^(1/2) + (4)^(1/2) - (3)^(1/2) = (4)^(1/2) - 1

therefore, the nth partial sum simplifies down to

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

and converges to infinity as n-> inf

b) (k=0) sum (inf) [x^(k)/(k!)]

looking for all x in R so it converges

I used the ratio test to get

| [(x)^(k+1)/(k+1)!] / [x^(k)/(k!)] | < 1

then I get -(k+1) < x < (k+1)

so if x is between those values, the series converges.

c) Converges by the comparison test


Hi, can someone let me know if I got these right? thanks!

[tiny-tim]

Hi squaremeplz! :smile:

(have a square-root: √ and a sigma: ∑ and an infinity: ∞ :wink:)

a) and c) are ok.
b) (k=0) sum (inf) [x^(k)/(k!)]

looking for all x in R so it converges

I used the ratio test to get

| [(x)^(k+1)/(k+1)!] / [x^(k)/(k!)] | < 1

then I get -(k+1) < x < (k+1)

so if x is between those values, the series converges.

But k goes up to ∞ :redface:

(and anyway you should be familiar with this series :wink:)

[squaremeplz]

ah, so it converges for all values of x?

[tiny-tim]

ah, so it converges for all values of x?

Yup! :biggrin:

And its value is … ex ? :wink: