Series Question

[oomjos]

The question asks to use a suitable gemotric series and the intergration or differentiation theorem to find the sum of the series.

So the first thing I tired to so was to shift the summation index like so

And then to write it like this so that I can see what my x is

Now that I know what m y x is I replace 1/2 with x because I will substitute it in later when evaluating the sum.

Now I know that a power series can be written as the following

So to get it in that form I differentiate my power series once. Not sure If I'm allowed to do this.

Now to fix the index of the summation again...

And now to write it in the correct form...

Now I know that.

But I have a different fucntion so I first integrated my function.

So this is the part when I realise I'm not on the right track because that doesn't look right.

Because If i evaluate the sum at 1/2 I get
1.16667

Any help would be appreciated.

Thanks

 

[Ishuda]

First - you did the derivative incorrectly. The index doesn't automatically change when you take the derivative. This lead to an incorrect integrand of x³/(1-x) when it should should have been x²/(1-x) which one can solve with partial fractions.

However, rather than try to integrate the x²/(1-x), let's do essentially the same thing but slightly differently: Consider
\(\displaystyle \Sigma_{n=3} \frac{x^n}{n}\)
Take the derivative (and yes you are allowed under certain circumstances which this series meets)
(\(\displaystyle \Sigma_{n=3} \frac{x^n}{n}\))'= \(\displaystyle \Sigma_{n=3} {x^{n-1}}\)
= \(\displaystyle \Sigma_{n=2} {x^{n}}\)
= -(1+x) + 1 + x + \(\displaystyle \Sigma_{n=2} {x^{n}}\)
= -(1+x) + \(\displaystyle \Sigma_{n=0} {x^{n}}\)
= -(1+x) + \(\displaystyle \frac {1} {1 - x}\)

Now integrate and
\(\displaystyle \Sigma_{n=3} \frac{1}{n\space 2^n}\) = \(\displaystyle \Sigma_{n=3} \frac{x^n}{n}|_0^{\frac{1}{2}} \)
\(\displaystyle = -(x + \frac{1}{2}\space x^2)|_0^{\frac{1}{2}} - ln(1 - x)|_0^{\frac{1}{2}}\)
= ln(2) - \(\displaystyle \frac{5}{8}\) [if I haven't made any dumb mistakes]

 

[Ishuda]

To get back to that derivative:
\(\displaystyle \Sigma_{n=3} \frac{x^n}{n} = \frac{x^3}{3} + \frac{x^4}{4} + \frac{x^5}{5} + ... \)
and
\(\displaystyle (\Sigma_{n=3} \frac{x^n}{n})' = \frac{3 x^2}{3} + \frac{4 x^3}{4} + \frac{5 x^4}{5} + ... = x^2 + x^3 + x^4 + ... = \Sigma_{n=3} x^{n-1}\)

Where you may have been put off is when you have a series for which a term has been dropped because it is zero after differentiation:
\(\displaystyle \Sigma_{n=0} x^n = 1 + x + x^2 + x^3 + ...\)
so that
\(\displaystyle (\Sigma_{n=0} x^n)' = 0 + 1 + x + x^2 + ...= \Sigma_{n=0} n\space x^{n-1}\)
or, dropping the zero term,
\(\displaystyle (\Sigma_{n=0} x^n)' = 1 + x + x^2 + ...= \Sigma_{n=1} n\space x^{n-1}\)

 

[oomjos]

That makes sense.

I think I misinterpreted the differentiation of a power series as you rightly pointed out.

I will try and do a few more of these using this procedure.

Thank you!