a logic-based one...

Prove that the sum of the infinite series 1+1-1+1-1+... is either 1 or 0.

<edited:>
the sum can be 1/2 too..

in fact, the sum is formally defined to be 1/2.
So...go ahead and prove it!

11 Answers

11
Devil ·

Take the sum uptil 'n' terms, ans comes out from whether n is odd or even.

1
Ritika ·

Can the sum not be 2?

11
Mani Pal Singh ·

good question by ritika
check he question again dude

1
neeti k ·

if sum is extend till n 4 +1...den
+1 +1 + 1.....n + -1 -1 -1....(.n-1)
= n - (n-1)
=1
..
if sum of -1 is extendd 2 n
den also
-1 -1 -1.......n + +1 +1...(n-1)
=-n+n-1
= -1
so sum will b alwys +/- 1.

1
Shreyan ·

oh yes....well noticed, ritika...

the series is 1-1+1-1+1-...

62
Lokesh Verma ·

What shreyan is probably hinting is to write this as S
then

S=1-S

thus, S=1/2

BTW if i remember correctly i had put a QOD on this funda... I could not mine it out! :(

1
Jagaran Chowdhury ·

1-1=0
1-1+1-1+1-1+.......
n(1-1)=0 if n is even
n(1-1)+1=1 if n is odd

is s=1/2 physically possible

1
Kalyan Pilla ·

@jagaran
S=1/2, is not possible, though the limiting value of S as the number of terms tends to ∞ is 1/2 if U solve it by the above methodology as told by Nishant sir

Though I particularly feel that the limit does not exist. But then if consider it as a series, then the answer is 1/2

The answer can also be obtained by this method.......

The given sequence is a GP with first term 1 and common ratio -1, which gives the sum as..

S=a/1-r =1/(1-(-1))
=1/2

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1
Jagaran Chowdhury ·

@kalyan how can you use formula for gp series here mod(common ratio) is not less than 1 .it is an infinitely oscilating series

62
Lokesh Verma ·

@Jagaran.. what you are saying is absoulutely true..

But this is an infinite series.. so waht will the sum be?

62
Lokesh Verma ·

@kalyan... your conclusion is wrong as given/ shown by Jagaran...

His post is correct

@Jagaran. (But when I am talking to you, I am arguing in the other direction ;)

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