Simplifying $\sum _{t=1}^{\infty }\:\frac{\left(1+g\right)^{ \lceil t/2\rceil}}{\left(1+k\right)^t}$

Question

how to simplify this expression so that I get an expression only?

when $t = 1, 2,.... $

$$y = \sum _{t=1}^{\infty }\:\frac{\left(1+g\right)^{ \lceil t/2\rceil}}{\left(1+k\right)^t}$$

I know that for t = 1, $\lceil \frac{t}{2}\rceil$ = 1

for t = 2, $\lceil \frac{2}{2}\rceil$ = 1

for t = 3, $\lceil \frac{3}{2}\rceil$ = 2

for t = 4, $\lceil \frac{4}{2}\rceil$ = 2

so in general $\lceil \frac{t}{2}\rceil = \frac{t}{2}$ if t is even, and $\lceil \frac{t}{2}\rceil = \frac{t+1}{2}$, if $t$ ia odd. And I can split this expression into two parts the even part and the odd part. The problem I am facing is I dont know how that affect the summation symbol, should I let $t = 2k$, and substitute $t = 2k$ for the even part and sub that into the summation and the t at the denominator?

EDIT: (continuing from @mag's suggestion)

$\sum _{t=1}^{\infty }\:\frac{\left(1+g\right)^{ \lceil \frac{t}{2}\rceil}}{\left(1+k\right)^t}$

$=\sum _{n=1}^{\infty }\:\frac{\left(1+g\right)^{\lceil n\rceil}}{\left(1+k\right)^{2n}}+\frac{\left(1+g\right)^{ \lceil\frac{2n-1}{2}\rceil}}{\left(1+k\right)^{2n-1}}$

$=\sum _{n=1}^{\infty }\:\frac{\left(1+g\right)^{n}}{\left(1+k\right)^{2n}} + \sum _{n=1}^{\infty }(\frac{\left(1+g\right)^{ n}}{\left(1+k\right)^{2n}} \times \frac{1}{\left(1+k\right)^{-1}})$

$= (1 + \frac{1}{\left(1+k\right)^{-1}}) \sum _{n=1}^{\infty }\:\frac{\left(1+g\right)^{n}}{\left(1+k\right)^{2n}} $

using the geometric sum to infinity formula: $$ \sum _{k=1}^{\infty }\: ar^{k-1} = \frac{a}{1-r}$$

and given that

$(\frac{1+g}{(1+k)^2})^{-1}\times(\frac{1+g}{(1+k)^2}) = 1$

$= (1 + \frac{1}{\left(1+k\right)^{-1}}) \sum _{n=1}^{\infty }\:\frac{\left(1+g\right)^{n}}{\left(1+k\right)^{2n}} \times (\frac{1+g}{(1+k)^2})^{-1}\times(\frac{1+g}{(1+k)^2})$

$= (1 + \frac{1}{\left(1+k\right)^{-1}}) \sum _{n=1}^{\infty }\:(\frac{\left(1+g\right)}{\left(1+k\right)^{2}})^n \times (\frac{1+g}{(1+k)^2})^{-1}\times(\frac{1+g}{(1+k)^2})$

$= (\frac{1+g}{(1+k)^2})(1 + \frac{1}{\left(1+k\right)^{-1}}) \sum _{k=1}^{\infty }\: 1 \times (\frac{1+g}{(1+k)^2})^{n-1}$

where $r = \frac{1+g}{(1+k)^2}$ and $a = 1$

and applying the formula:

$= (\frac{1+g}{(1+k)^2})(1 + \frac{1}{\left(1+k\right)^{-1}}) \times \frac{1}{1- (\frac{1+g}{(1+k)^2})}$

I can try to simplify even more but just wanted to make sure I am on the right track, especially in terms of applying the infinite geometric sum formula.

There is also another part of the question which is that what is the condition of g so that $y$ is finite. In this case, since $|r| < 1$

$r = \frac{1+g}{(1+k)^2} < 1$

$g < k(k+2)$

Answer 1

Yes, what you're describing is (almost) correct. Just be carefull as $k$ is already a variable in your calculation you have to substitute with a "free" variable like $t=2n$ for the even part and $t=2n-1$ for the odd part. $$\sum _{t=1}^{\infty }\:\frac{\left(1+g\right)^{ \lceil \frac{t}{2}\rceil}}{\left(1+k\right)^t}=\sum _{n=1}^{\infty }\:\frac{\left(1+g\right)^{\lceil n\rceil}}{\left(1+k\right)^{2n}}+\frac{\left(1+g\right)^{ \lceil\frac{2n-1}{2}\rceil}}{\left(1+k\right)^{2n-1}}$$ By using $\lceil\frac{2n-1}{2}\rceil=n$ you have simplified the ceil symbol.

Answer 2

You can rewrite the series as

$$ \sum_{r=0}^\infty\left({(1+g)^r\over(1+k)^{2r+1}}+{(1+g)^{r+1}\over(1+k)^{2r+2}}\right) $$

Which is just taking two terms at once

Now you can split the sums into two infinite g.PS which you should be able solve easily