Determine-1-2-1-2-3-1-3-

Question Number 3022 by Rasheed Soomro last updated on 03/Dec/15

Determine

1 + 2^{\frac{1}{2}} + 3^{\frac{1}{3}} + \dots

Commented by prakash jain last updated on 04/Dec/15

y = x^{1 / x}

\ln y = \frac{\ln x}{x}

\lim_{x \to \infty} \frac{\ln x}{x} = \lim_{x \to \infty} \frac{1}{x} = 0

\lim_{x \to \infty} {(x)}^{1 / x} = 1

The given series does not converge .

Commented by Filup last updated on 03/Dec/15

\underset{i = 1}{\sum^{\infty}} i^{\frac{1}{i}} \Rightarrow \underset{i = 1}{\sum^{\infty}} i^{i^{- 1}} \Rightarrow \underset{i = 1}{\sum^{\infty}}^{i} \sqrt{i}

\lim_{s \to 0} (\underset{i = 1}{\sum^{\infty}} i^{\frac{1}{i} - s}) = ζ (- \frac{1}{i}) note : i \neq \sqrt{- 1}

Commented by Rasheed Soomro last updated on 03/Dec/15

D {i d n}^{'} t u n d e r s t a n d . T r y a g a i n p l e a s e!

Commented by Filup last updated on 04/Dec/15

Because the given serious converges

to value 1 (i^{- i} n o t S), the series at i \to \infty

is about \dots 1 + 1 + 1 + 1 + \dots

with each value getting closer to 1 .

This makes the series not limit to a single

value .

Hence showing evidence that ζ (- \frac{1}{i}) = \infty

and S_{2} in some cases

Commented by prakash jain last updated on 04/Dec/15

A series S = \underset{n = 1}{\sum^{\infty}} a_{n} converges only \lim_{n \to \infty} a_{n} = 0 .

This is necessary condition .

Commented by Rasheed Soomro last updated on 04/Dec/15

T h a n k S!

Answered by Filup last updated on 03/Dec/15

for S = ζ (- s), S \to \infty

There maybe be an analytical

approach such as ζ (- 1) = Σ N^{+} = - \frac{1}{12}

My unproven solution would be that

for S = \underset{i = 1}{\sum^{\infty}} i^{\frac{1}{i}} = ζ (- \frac{1}{i})

a s i \to \infty, S \to \infty

So, I would suggest that

S = {\infty, S_{2}}

Where S_{2} is an analytical solution

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