k=1 sigma k tends to infinity ((1/2^k)-(1/2(k+1))) how can calculate the sum of this series..?

mlehuzzah | Certified Educator

First consider the first piece:

`sum_(k=1)^(oo) (1)/(2^k) = (1)/(2)+(1)/(4)+(1)/(8)+...`

This is a geometric series.  The ratio, r, is `(1)/(2)`, which means that each term is `(1)/(2)` times the previous term.  The first term, a, is what you get when you plug `k=1` into `(1)/(2^k)`, which is `(1)/(2)`

So:

`r=(1)/(2)`
`a=(1)/(2)`

and we plug these into the formula for an infinite geometric series, with `|r|<1`

`a (1)/(1-r) = (1)/(2)*(1)/(1-(1)/(2)) = (1)/(2)*2=1`

So

`sum_(k=1)^(oo) (1)/(2^k) = (1)/(2)+(1)/(4)+(1)/(8)+...= 1`

This is FINITE.  That means we can separate it out, and examine the second piece separately.

More precisely:

`a_k-b_k=c_k`

If `sum a_k` is finite, then:

If `sum b_k` is divergent, then `sum c_k` is divergent, and
If `sum b_k` is convergent, then `sum c_k` is convergent.

It turns out that `sum_(k=1)^(oo) (1)/(2(k+1))` is divergent.
Its sum is infinity.

To see this, we use the fact that the harmonic series is divergent:
Harmonic series:

`1+(1)/(2)+(1)/(3)+(1)/(4)+...`

Our series is:

`(1)/(2)[(1)/(2)+(1)/(3)+(1)/(4)+...]`

`=(1)/(2)["harmonic" - 1]`

If you take an infinite series, and subtract a finite number, the result is still infinite.  If you take an infinite series, and multiply by a nonzero finite number, the result is still infinite.  Thus, our series is divergent.

Thus:

`sum_(k=1)^(oo) (1)/(2^k) = 1`

`sum_(k=1)^(oo) (1)/(2(k+1))` is divergent.

So

`sum_(k=1)^(oo) (1)/(2^k) - (1)/(2(k+1))` is divergent.

PS:

To see WHY the harmonic series is divergent, consider:

`1+(1)/(2)+(1)/(3)+(1)/(4)+(1)/(5)+(1)/(6)+(1)/(7)+(1)/(8)+(1)/(9)+(1)/(10)+...`

`1+(1)/(2)+(1)/(4)+(1)/(4)+(1)/(8)+(1)/(8)+(1)/(8)+(1)/(8)+(1)/(16)+(1)/(16)+...`

The top series will have a larger sum, because all of the terms are greater than or equal to the corresponding terms in the second series

But consider the sum of the second series:

`1+(1)/(2)+(1)/(4)+(1)/(4)+(1)/(8)+(1)/(8)+(1)/(8)+(1)/(8)+(1)/(16)+(1)/(16)+...`

`=1+(1)/(2)+2*(1)/(4)+4*(1)/(8)+8*(1)/(16)+...`

`=1+(1)/(2)+(1)/(2)+(1)/(2)+...`

If you add `(1)/(2)` to itself for forever, you don't get something finite.

So, the SMALLER of our series is infinite, and so the larger one (the harmonic series) is also infinite.