Homework: Investigating Infinite Series Convergence

[squaremeplz]

Homework Statement

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.

Homework Equations

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 squaremeplease!

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

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 ∞

(and anyway you should be familiar with this series )

 

[squaremeplz]

ah, so it converges for all values of x?

 

[tiny-tim]

ah, so it converges for all values of x?

Yup!

And its value is …

Spoiler

e^x

? ​