1. ## convergence series

Hey All,

I have two questions on series which I have done and just need someone to check them for me. Thank you

1) find range of value of x for the convergence series:
x + 2^4.x^2/2! + 3^4.x^3/3! + 4^4.x^4/4! + ...

so:
Un = n^4x^n/n!

Un + 1 = (n+1)^4x^{n+1}/(n+1)!

(Un + 1)/(Un) = (n+1)^4x^{n+1}/(n+1)! . n!/n^4x^n

(x+y)^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4 therefore:
(n+1)^4 = n^4 + 4n^3 +6n^2 + 4n + 1

(Un + 1)/(Un) = (n^4 + 4n^3 +6n^2 + 4n + 1)x^n.x^1/n! + 1! . n!/n^4x^n

Simplifying:
(Un + 1)/(Un) = (4n^3 +6n^2 + 4n + 1)x^1/1!

Lim n→∞ (Un + 1)/(Un) = (4 + 6/n + 4/n^2 + 1/n^3)x/n^3 / 1!/n^3

= (4 + 0 + 0 + 0)0 / 0 = 0

converges

2) investigate convergence of series
1/1.2 + 1/2.2^2 + 1/3.2^3 + 1/4.2^4 + ...

so:
Un = 1 / n2^n
Un + 1 = 1 / (n+1)2^{n+1}

(Un + 1)/(Un) = 1 / (n+1)2^n.2^1 . n2^n/1

(Un + 1)/(Un) = n2^n / (n+1)2^n.2^1

(Un + 1)/(Un) = n / (n+1)2

Lim n→∞ (Un + 1)/(Un) = 0 / (0 +1)2 = 0

converges

Your algebra is a bit off . . . and your limits are way off.

1) Find the inverval of convergence for:

. . x + (2^
4·x^2)/2! + (3^4·x^3)/3! + (4^4·x^4)/4! + . . .
Un+1 . . . . (n+1)^4·x^{n+1} . . .n! . . . . . .(n+1)^4 .x^{n+1} . .n!
------ . = . -------------------·----------- . = . ---------·----------·--------
. U
n . . . . . . . . (n+1)! . . . .n^4·x^n . . . . . n^4 . . . .x^n . .(n+1)!

. . . . .(n+1)^
4 . .x . . . . .(n+1)³ . . . . . n³ + 3n² + 3n + 1
. . = . ----------·----- . = . --------·x . = . ----------------------·x
. . . . . . n^
4 . . n+1 . . . . . n^4 . . . . . . . . . . n^4

. . . . . . . . . . . . . . . . . . . . . . .1/n + 3/n² + 3/n³ + 1/n^
4
Divide top and bottom by n^
4: . -------------------------------·x
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

Then: .lim (1/n + 3/n² + 3/n³ + 1/n^
4)·x . = . 0
. . . . n→∞

Therefore, the series converges for all values of x.

2) investigate convergence of the series:
. . 1/(1·2) + 1/(2·2^2) + 1/(3·2^3) + 1/(4·2^4) + . . .

U
n+1 . . . . . . . . 1 . . - . . .n·2^n . . - . . n - . 1
------ . = . -----------------·-------- . = . -------·--
. U
n . . . . .(n+1)·2^{n+1} . . 1 . . . . . .n + 1 .2

. . . . . . . . . . . . . . . . . . . . . . . 1 . . . .1
Divide top and bottom by n: . --------- · --
. . . . . . . . . . . . . . . . . . . . . 1 + 1/n . 2

. . . . . . . . . . .1 . - - .1 . . . . 1
Then: .lim .---------- · -- . = . --
. . . . x→∞ .1 + 1/n . .2 . . . . 2

Therefore, the series converges.