Question

∑∞ n=1 (n^2n)/(1+n)^(3n)
Does it converge or diverge?
What method did you use?
PLEASE HELP

Answer 1

Answer:

Given the series,

∑ ∞ n = 1 − 4 ( − 1 / 2 ) n − 1

I think the series is summation from n = 1 to ∞ of -4(-1/2)^(n-1)

So,

∑ − 4 ( − ½ )^(n − 1). From n = 1 to ∞

There are different types of test to show if a series converges or diverges

So, using Ratio test

Lim n → ∞ (a_n+1 / a_n)

Lim n → ∞ (-4(-1/ 2)^(n+1-1) / -4(-1/2)^(n-1))

Lim n → ∞ ((-4(-1/2)^(n) / -4(-1/2)^(n-1))

Lim n → ∞ (-1/2)ⁿ / (-1/2)^(n-1)

Lim n→ ∞ (-1/2)^(n-n+1)

Lim n→ ∞ (-1/2)^1 = -1/2

Since the limit is less than 0, then, the series converge...

Sum to infinity

Using geometric progression formula

S∞ = a / 1 - r

Where

a is first term

r is common ratio

So, first term is

a_1 = -4(-½)^1-1 = -4(-½)^0 = -4 × 1

a_1 = -4

Common ratio r = a_2 / a_1

a_2 = 4(-½)^2-1 = -4(-½)^1 = -4 × -½ = 2

a_2 = 2

Then,

r = a_2 / a_1 = 2 / -4 = -½

S∞ = -4 / 1--½

S∞ = -4 / 1 + ½

S∞ = -4 / 3/2 = -4 × 2 / 3

S∞ = -8 / 3 = -2⅔

The sum to infinity is -2.67 or -2⅔

Step-by-step explanation: PHEW THAT TOOK A WHILE LOL IM A FAST TYPER

Answer 2

It looks like the series is

$\displaystyle \sum_{n=1}^\infty \frac{n^{2n}}{(1+n)^{3n}} = \sum_{n=1}^\infty \left(\frac{n^2}{(1+n)^3}\right)^n$

Checking for convergence is most easily done with the (Cauchy) root test for convergence: the infinite series

$\displaystyle \sum_{n=1}^\infty a_n$

converges absolutely if the limit

$\displaystyle \lim_{n\to\infty} \sqrt[n]{|a_n|} < 1$

Here we have

$\displaystyle \lim_{n\to\infty} \sqrt[n]{\left|\left(\frac{n^2}{(1+n)^3}\right)^n\right|} = \lim_{n\to\infty} \frac{n^2}{(1+n)^3} = 0 < 1$

so the given series converges.