Question

Test the convergence of infinite series of natural numbers. (NOTE: its a self attempted question, a technique used by socratians earlier)

Answer 1

So, as we increase value of n the sum diverges more and more but sum of first infinite natural numbers converges on the negative side of the axis at −1/12 , everybody could think that the sum would tend to be infinite but infact it doesn't.

[Diverges and converges here mean to convey that if you plot a xy graph of given series with number of terms on x axis and sum of y axis , as you increase the terms; y coordinate tends to become more and more larger butif we continue to plot the graph till infinity, which though is impossible, we'll get y-coordinate as −1/12]

Here's the proof

Consider

Sn=1−1+1−1+1−1.......

If we add some brackets we notice that sum of this series would be

Sn=(1−1)+(1−1)+(1−1)+.....

Sn=0+0+0+0....=0

But if we just change position of brackets

Sn=1−(1−1)−(1−1)−(1−1)..

Sn=1−0−0−0−0−0....=1

So, we take the average<br>

Sn=1/2

Alternate proof, this is better one.

S=1−1+1−1...

1−S=1−(1−1+1−1+1−1+1−1......)

1−S=1−1+1−1+1.....

therefore, 1−S=S

therefore, 1=2S

S=1/2

Now consider

Sn&apos;=1−2+3−4+5−6+7−8....

Just add Sn' to itself<br>

Sn'=1−2+3−4+5−6...

Sn'= 1−2+3−4+5...

(we have shifted the numbers to little right, because the sum is like 1−(2−1)+(3−2).... for a finite series this method is invalid, here we are adding nth term with n+1th term with whatever signs they have before them.)

2Sn=1−1+1−1+1−1...=Sn=1/2

Sn'=1/4

Now, our main series

Sn"=1+2+3+4+5...

Subtract Sn' from this series.

Sn"= 1+2+3+4+5+6....

−Sn'=−1+2−3+4−5+6..

Sn"−Sn'=4+8+12+16...

Sn"−1/4=4(1+2+3+4...)

Sn"=4Sn"+1/4

Sn"−4Sn"=1/4<

−3Sn"=1/4<br>

Sn"=1/4⋅1/(−3)=−1/12