Test-for-convergence-1-n-10-2-ln-lnn-nlnn-2-n-2-1-n-lnn-p-two-cases-of-p-to-look-at-3-n-2-1-n-n-lnn-4-n-1-10-n-n-2n-1-5-n-1-

Question Number 3564 by Yozzii last updated on 15/Dec/15

T e s t f o r c o n v e r g e n c e :

(1) \underset{n = 10}{\sum^{\infty}} \frac{2^{\ln (\ln n)}}{n \ln n}

(2) \underset{n = 2}{\sum^{\infty}} \frac{1}{n {(\ln n)}^{p}} (two cases of p to look at)

(3) \underset{n = 2}{\sum^{\infty}} \frac{{(- 1)}^{n} \sqrt{n}}{\ln n}

(4) \underset{n = 1}{\sum^{\infty}} \frac{10^{n} n}{(2 n + 1)!}

(5) \underset{n = 1}{\sum^{\infty}} \frac{n!}{n^{n}}

T h i s p o s t i s m y a t t e m p t t o a d h e r e t o

t h e c o m m o n a g r e e m e n t o n i n f i n i t e

s e r i e s s t u d y i n g . H a v e f u n!

Commented by prakash jain last updated on 15/Dec/15

Find sum or only check for convergence ?

Commented by Yozzii last updated on 15/Dec/15

D o a s y o u p l e a s e, b u t I o n l y a s k e d

a b o u t c o n v e r g e n c e . I f a q u e s t i o n

s p a r k s f u r t h e r q u e s t i o n s t h e n f a n t a s t i c!

Answered by prakash jain last updated on 15/Dec/15

5 . a_{n} = \frac{n!}{n^{n}}, a_{n + 1} = \frac{(n + 1)!}{{(n + 1)}^{n + 1}}

\frac{a_{n + 1}}{a_{n}} = \frac{(n + 1)! n^{n}}{n! {(n + 1)}^{n + 1}} = \frac{(n + 1) n^{n}}{(n + 1) {(n + 1)}^{n}} = {(\frac{n}{1 + n})}^{n}

\lim_{n \to \infty} {(\frac{n}{1 + n})}^{n} = \lim_{x \to 0} {(\frac{1 / x}{1 + 1 / x})}^{1 / x} = \lim_{x \to 0} {(\frac{1}{1 + x})}^{1 / x}

y = {(\frac{1}{1 + x})}^{1 / x} \Rightarrow \ln y = - \frac{\ln (1 + x)}{x}

\lim_{x \to 0} \ln y = \lim_{x \to 0} - \frac{\frac{1}{1 + x}}{1} = - 1

\lim_{x \to 0} {(\frac{1}{1 + x})}^{1 / x} = \frac{1}{e}

\frac{1}{e} < 1, hence by ratio test series converges .

Answered by prakash jain last updated on 15/Dec/15

4 . a_{n} = \frac{10^{n} n}{(2 n + 1)!} \Rightarrow \frac{a_{n + 1}}{a_{n}} = \frac{10 (n + 1)}{n (2 n + 2) (2 n + 3)}

\lim_{n \to \infty} \frac{10 (n + 1)}{n (2 n + 2) (2 n + 3)} = 0

series converges

Answered by prakash jain last updated on 15/Dec/15

3 . a_{n} = \frac{{(- 1)}^{n} \sqrt{n}}{\ln n}

Ratio test, r_{n} = \frac{\sqrt{n + 1}}{\ln (n + 1)} \times \frac{\ln n}{\sqrt{n}}

= \sqrt{1 + \frac{1}{n}} \times \frac{\ln n}{\ln n + 1}

\lim_{n \to 0} r_{n} = 1 (inconclusive)

\lim_{n \to \infty} ∣ a_{n} ∣= \lim_{n \to \infty} \frac{\sqrt{n}}{\ln n} = \frac{\infty}{\infty}

\lim_{n \to \infty} ∣ a_{n} ∣= \lim_{n \to \infty} \frac{\sqrt{n}}{\ln n} = \lim_{n \to \infty} \frac{n}{2 \sqrt{n}} = \infty

series diverges .