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Sequences/Series Question

nacho

Active member
Sep 10, 2013
156
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.
 

Attachments

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
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.
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 $\left|\dfrac{i}{3}\right| = \dfrac{|i|}{3} = \dfrac{1}{3} < 1$.

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

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

nacho

Active member
Sep 10, 2013
156
Thanks 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 $\frac{1}{1-\frac{i}{3}}$
 

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
Thanks 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 $\frac{1}{1-\frac{i}{3}}$
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.
 

nacho

Active member
Sep 10, 2013
156
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.
this i understood,
what i did not know was how they arrived at the sum being equal to
$\frac{1}{1-\dfrac{i}{3}}$
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Use that power series

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

nacho

Active member
Sep 10, 2013
156