0
$\begingroup$

$\dfrac{1}{z+3i}$ can be interpreted as the sum of the geometric series $\displaystyle\dfrac{1}{6i}\sum_{k=0}^{\infty}\left(\frac{z-3i}{6i}\right)^n$ this can be obtained by writing: $\dfrac{1}{z+3i}=\dfrac{1}{6i+(z-3i)}=\dfrac{1}{6i}\dfrac{1}{1+\dfrac{1}{6i}(z-3i)}$ now the book I'm reading says that this is equal to the series I said at the beginning. But how is it possible if the sum of a geometric series is in the form of $\dfrac{1}{1-q}$?

Edit: I am sure there is convergence of the geometric sum because $|z-3i|<5$

$\endgroup$
  • $\begingroup$ It looks to me like the sum should be ${1\over 3i-z}$ Is that what you are asking? (I'm assuming that $n$ and $k$ are supposed to be the same letter in the expression for the sum.) Also, why do you say, $|z-3i|<5, z\in\mathbb{C}?$ $\endgroup$ – saulspatz Nov 23 '18 at 16:40
  • $\begingroup$ Because I need that the centre of the geometric series is $3i$ $\endgroup$ – pter26 Nov 23 '18 at 17:34
  • $\begingroup$ For the series to converge $|z-3i|<6$. $\endgroup$ – Yadati Kiran Nov 23 '18 at 17:36
  • $\begingroup$ But a priori |z-3i|<5 in my problem $\endgroup$ – pter26 Nov 23 '18 at 17:38
0
$\begingroup$

Probably I don't fully understand your question. But let's make a try.

Consider the following series:

$$S = q + qx + qx^2 + qx^3 + \ldots+qx^{n-1} $$

Now let's multiply the whole series by $x$:

$$Sx = qx + qx^2 + qx^3 + qx^4 + \ldots+qx^n$$

Now let's subtract the second from the first:

$$S - Sx = (q + qx + qx^2 + qx^3 + \ldots+qx^{n-1}) - (qx + qx^2 + qx^3 + qx^4 + \ldots+qx^n)$$

We can easily show that only few terms survive at the right side, that is,

$$S(1 - x) = q - qx^n$$

Whence

$$S(1-x) = q(1-x^n)$$

$$S = q\frac{1-x^n}{1-x}$$

As $n$ goes to infinity, the absolute value of $x$ must be less than $1$ for the series to converge, the sum then becomes

$$q + qx + qx^2 + qx^3 + \ldots = \sum_{k = 0}^{+\infty} qx^k = \frac{q}{1-x} ~~~~~~~ |x| < 1$$

And for $q = 1$ we just get

$$\frac{1}{1-x}$$

As wanted.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.