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

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mlehuzzah | Student, Graduate | (Level 1) Associate Educator

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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)`



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`


`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:


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:


Our series is:


`=(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.


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

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


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


To see WHY the harmonic series is divergent, consider:



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:




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.

For more info, see:

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