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- Thread starter nacho
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- Jan 26, 2012

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Don't let the $i$'s get to you! For instance, in 1(a), $\displaystyle\sum_{j=0}^{\infty} \left(\frac{i}{3}\right)^j$ is geometric since $\left|\dfrac{i}{3}\right| = \dfrac{|i|}{3} = \dfrac{1}{3} < 1$.Please refer to the attached image.

How do I do these questions, In particular

1a, 1c and 1f.

Could anyone give me a hint to get me started with either of these?

Thanks.

Note that you can rewrite 1(c) as $\displaystyle\sum_{j=0}^{\infty}\left(-\frac{2}{3}\right)^j$.

Note that 1(f) is a telescoping sum; with that said, we see that

\[\sum_{j=0}^{\infty} \left[\frac{1}{j+2} - \frac{1}{j+1}\right] = \lim_{n\to\infty} \sum_{j=0}^n \left[\frac{1}{j+2} - \frac{1}{j+1}\right] = \ldots\]

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- Jan 26, 2012

- 995

Note thatThanks for that.

Although, for

1a) it says that the sum of the series is

$\frac{1}{1-\frac{i}{3}}$ $ = $ $\frac{9+3i}{10}$

Why is this? I am unsure how they get to the first term of $\frac{1}{1-\frac{i}{3}}$

\[\frac{1}{1-\dfrac{i}{3}} = \frac{3}{3\left(1-\dfrac{i}{3}\right)} = \frac{3}{3-i}\]

Now conjugate and you'll get the desired result.

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this i understood,Note that

\[\frac{1}{1-\dfrac{i}{3}} = \frac{3}{3\left(1-\dfrac{i}{3}\right)} = \frac{3}{3-i}\]

Now conjugate and you'll get the desired result.

what i did not know was how they arrived at the sum being equal to

$\frac{1}{1-\dfrac{i}{3}}$

- Jan 17, 2013

- 1,667

Use that power series

\(\displaystyle \frac{1}{1-z} = \sum_{k=0}^{\infty}z^k \,\,\,\,\,\,\, |z|<1\)

\(\displaystyle \frac{1}{1-z} = \sum_{k=0}^{\infty}z^k \,\,\,\,\,\,\, |z|<1\)

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thanks!Use that power series

\(\displaystyle \frac{1}{1-z} = \sum_{k=0}^{\infty}z^k \,\,\,\,\,\,\, |z|<1\)