REACH THE IMPOSSIBLE DREAM
In the situation in which somehow, in some fashion, by some trick - this 2 different series (see below) had a $$\color{red}{\bf FINITE\quad Total \quad Sum}$$, what would such Total Sums be ?
First series
$$\sum_{k=1}^\infty \frac{1}{k}$$
Second series
$$\sum_{k=1}^\infty (-1)^{k+1}$$

What would the sums be ?

Question

REACH THE IMPOSSIBLE DREAM
In the situation in which somehow, in some fashion, by some trick - this 2 different series (see below) had a $$\color{red}{\bf FINITE\quad Total \quad Sum}$$, what would such Total Sums be ?
First series
$$\sum_{k=1}^\infty \frac{1}{k}$$
Second series
$$\sum_{k=1}^\infty (-1)^{k+1}$$

What would the sums be ?

anonymous · Answer

Solutions Exists.

anonymous · Answer

But this is infinite @sauravshakya

anonymous · Answer

id like to know what is solution...so watching :)

anonymous · Answer

ME too.

anonymous · Answer

Well there was a MAN from your part of the world... VERY famous.... mystical mathematician... numbers were his personal friends
1) Who was HE ?
2) How does he help you in that ?

anonymous · Answer

CALCULATOR

anonymous · Answer

@estudier

anonymous · Answer

http://en.wikipedia.org/wiki/Srinivasa_Ramanujan

anonymous · Answer

Are u sure the question is correct....... i think it is 1/k instead of 1/n

anonymous · Answer

Cesaro/Grandi

anonymous · Answer

Abel

anonymous · Answer

You get nothing @estudier  . You are the destructor of PLEASURES

anonymous · Answer

THE SECOND ONE IS 0

anonymous · Answer

1/2

anonymous · Answer

I think for first one it is greater than 2

anonymous · Answer

Divergent series upset the calculs addicts....

anonymous · Answer

$$  Lao\,\, Tzu\,\,\,says\,\,\, \text{There is no such thing as divergent series} $$

anonymous · Answer

No, that was Sum Tzu..

anonymous · Answer

Rubbish estudier, tat guy was only a stupid general

anonymous · Answer

Um, you didn't get my little joke....(sum not sun)

anonymous · Answer

Yes I got it - but the mind was faster then the hand.

anonymous · Answer

WELL, if @Mikael THE ONE WHO KNOWS EVERYTHING HAS ALREADY SOLVED THEN WHY SHOULD I DO IT......

anonymous · Answer

Look Abel summation or Cesaro summation - AND PLEASE SOLVE IT here. PLease!

anonymous · Answer

Grandi series

anonymous · Answer

The real understanding comes from Abel summation !

anonymous · Answer

Cesaro

anonymous · Answer

I thought everyone knew that one.....

anonymous · Answer

im back... lol

anonymous · Answer

@mukushla  is brilliant in solving problem.......... @Mikael

anonymous · Answer

One of best talent I have seen.

anonymous · Answer

But you didn't look up Cesaro or Abel summation

anonymous · Answer

1+1+2+5+14+42+...= 1/2 - i sqrt3/2

anonymous · Answer

@sauravshakya @mukushla this sneaky French guy (cannot say his name) is stealing your chance at glory

anonymous · Answer

which glory ?

anonymous · Answer

1 +2 + 4 + 8 +16 +....=-1

anonymous · Answer

Somming DIVERGENT SERIES

anonymous · Answer

thats not real :)

anonymous · Answer

Them Catalan numbers again..........:-)

anonymous · Answer

Neither are imaginary numbers......

anonymous · Answer

Yep the Catalan numbers are back @estudier

anonymous · Answer

http://en.wikipedia.org/wiki/Ces%C3%A0ro_summation http://www.encyclopediaofmath.org/index.php/Abel_summation_method

anonymous · Answer

@sauravshakya  please read and apply to these two series

anonymous · Answer

or Mrs. @satellite73 or @experimentX ?

experimentx · Answer

lol ... you should be putting Mrs.

anonymous · Answer

second one is in the link you sent

anonymous · Answer

I meant- gentlemen ?

anonymous · Answer

a quick google search turns up euler mascheroni constant for harmonic series, which also seem so be 
$$\lim_{n\to \infty}\sum_{k=1}^nk-\ln(n)$$

experimentx · Answer

$$ \sum_{k=1}^\infty (-1)^{k+1}  = 1 - 1 +1 - 1 + 1 ... \
$$
some guy had me shown some trick that it converges to 1/2 ... don't remember it though.

anonymous · Answer

that is NOT A SOLUTION!

anonymous · Answer

http://en.wikipedia.org/wiki/Ces%C3%A0ro_summation http://www.encyclopediaofmath.org/index.php/Abel_summation_method

anonymous · Answer

As you know $$  \ln(n) \to \infty$$

anonymous · Answer

So Mascheroni is NOT the answer

experimentx · Answer

$$
\text {Let } 1 - 1 +1 - 1 + 1 -1 + ...  = a \
1 - (1 - 1 + 1 - 1 ...) = 1 - a = a \implies  2a = 1 \implies a = {1 \over  2 }$$

experimentx · Answer

this is false ... . of course :D

experimentx · Answer

By simple trick 
$$ \sum_{k=1}^\infty \frac{1}{k} $$
it can be shown to be divergent ... for I don't remember any sleight of hand on this. Let's think for a while.

anonymous · Answer

http://en.wikipedia.org/wiki/Ces%C3%A0ro_summation http://www.encyclopediaofmath.org/index.php/Abel_summation_method

anonymous · Answer

FOR Ramanujan's  sake READ !

anonymous · Answer

Messrs. @experimentX @satellite73 @sauravshakya

anonymous · Answer

1-1+2-6+24-120+......

anonymous · Answer

That will only work with Abel x> 5

anonymous · Answer

Too divergent - works with Double-Abel

anonymous · Answer

1-2+3-4+....

anonymous · Answer

Cesaro is enough for that

anonymous · Answer

The nasty one is 0.6 or thereabouts.....

anonymous · Answer

The nasty one is 0.6 or thereabouts.....

anonymous · Answer

Hey people - how can one cause you to read ?

anonymous · Answer

http://en.wikipedia.org/wiki/Borel_summation

experimentx · Answer

Ah ... i was working on project euler problems.

experimentx · Answer

Man I flunked in life for the first time on on analysis.

anonymous · Answer

You need divergent sum therapy....:-)

experimentx · Answer

How do you plan to sum with Abel summation?

anonymous · Answer

$$\lim_{x\rightarrow 1-0} \sum_{k=0}^\infty a_k x^k = S.$$

anonymous · Answer

$$\Huge \lim_{x\rightarrow 1-0} \sum_{k=0}^\infty a_k x^k = S.$$

experimentx · Answer

go on ...

anonymous · Answer

."..it is no more doubtful that the sum of this series 1−2+3−4+5 + etc. is 1⁄4; since it arises from the expansion of the formula 1⁄(1+1)2, whose value is incontestably 1⁄4." Euler

experimentx · Answer

interesting ... how do you do that?

experimentx · Answer

1⁄(1+1)2 = 1/2 ( 1 - 1 + 1 - 1 + 1 - 1 ... )

anonymous · Answer

half of half is a quarter......!

experimentx · Answer

how did you manage to get n's in coefficient.

experimentx · Answer

let's tyr this $$ {1 \over (1 + x)^2} = $$ http://www.wolframalpha.com/input/?i=expand+1%2F%281%2Bx%29^2 Idiot me!!

anonymous · Answer

http://en.wikipedia.org/wiki/1_%E2%88%92_2_%2B_3_%E2%88%92_4_%2B_%C2%B7_%C2%B7_%C2%B7

experimentx · Answer

should have noted.

anonymous · Answer

Scroll down to Euler and Borel

anonymous · Answer

Sorry, to Abel Summation....

experimentx · Answer

Interesting!!

experimentx · Answer

encountering this material for the first time man!!

anonymous · Answer

Good So just dive in and understand it "like a boss"...

experimentx · Answer

Hhaha i wish i could understand it "like a boss" ... I a bit slow. especially on algebra and analysis.

anonymous · Answer

Plug the sequence into the expression and try to find the limit

anonymous · Answer

The harmonic series is easily summed by Cesaro Summation

anonymous · Answer

Cesaro. Never heard of it. Thanks for bringing it accross.

experimentx · Answer

$$ \int {1 \over 1 -x}dx = \sum_{n=1}^\infty {x^n \over n }$$
I wonder if this is useful.

experimentx · Answer

the partial sums are
$$ {1}, {3 \over 4}, {11 \over 18}, {25 \over 48}, ...$$

experimentx · Answer

this seems to sonverge to zero as n->inf

experimentx · Answer

this is bugging me
$$ \lim_{m\rightarrow \infty} {1 \over m }\sum_{n=1}^m {1 \over n } \sum_{k=1}^n {1 \over k}$$

experimentx · Answer

I wonder if i can use this http://en.wikipedia.org/wiki/Faulhaber%27s_formula