MHB Convergence of a geometric series

[Manfred1999]

Hi everyone,

I am generally familiar with convergent series. However, in one economics paper (Becker&Tomes 1979), I found the following that confuses me:$$\sum_{j=0}^{k} \beta^{j} h^{k-j} = \beta^{k}(k+1)\quad \text{if} \quad\beta =h$$

however,

$$\sum_{j=0}^{k} \beta^{j} j^{k-j} = \frac{\beta^{k+1}-h^{k+1}}{\beta-h}\text{if} \quad\beta \ne h$$In short, I do not understand how they derived at these convergences. Who has any idea that are they referring to? The text itself does not provide more information. And despite having consulted those maths references I can access, I could not find an answer.

Thank you very much for any hint.

Man

PS
I hope the formulae are depicted properly

 

[Euge]

Hi Manfred1999,

I corrected the formatting of the equations you presented, and I hope what you see is what you intended to write. But, in the first equation, you did not complete the condition on $\beta$; $\beta$ equals what exactly?

 

[Manfred1999]

Hi Euge,

Thanks so much for doing this. Much appreciated. I think now I know what I did do wrong. I forgot the $ signs. Also, I did not notice that the conditions where not depicted properly. I hope it is clear now. Do you have any idea how do prove the convergence of the series in both cases? Maybe I should that k goes to \infty!
Man

Hi Manfred1999,

I corrected the formatting of the equations you presented, and I hope what you see is what you intended to write. But, in the first equation, you did not complete the condition on $\beta$; $\beta$ equals what exactly?

 

[Manfred1999]

I am an idiot! I just solved the case case. The one where beta goes to infinity! I am an idiot. I am happy to jot down my steps if that helps!

Hi Euge,

Thanks so much for doing this. Much appreciated. I think now I know what I did do wrong. I forgot the $ signs. Also, I did not notice that the conditions where not depicted properly. I hope it is clear now. Do you have any idea how do prove the convergence of the series in both cases? Maybe I should that k goes to \infty!
Man

 

[Euge]

$$\sum_{j=0}^{k} \beta^{j} j^{k-j} = \frac{\beta^{k+1}-h^{k+1}}{\beta-h}\quad\text{if} \quad\beta \ne h$$

The expressions $j^{k-j}$ should be $h^{k-j}$. To prove the corrected formula, perform synthetic division. Alternatively, multiply the sum $\sum\limits_{j = 0}^n \beta^j h^{k-j}$ by $\beta - h$; this gives

$$(\beta - h)\sum_{j = 0}^k \beta^j h^{k-j} = \sum_{j = 0}^k (\beta - h)\beta^j h^{k-j} = \sum_{j = 0}^k [\beta^{j+1}h^{k-j} - \beta^j h^{k-(j-1)}] = \sum_{j = 0}^k (x_{j+1} - x_j),$$

where $x_j = \beta^{j}h^{k-(j-1)}$. The sum you see at end is known as a telescoping sum. When you expand the sum, the only term that "survive" are $x_{n+1}$ and $x_0$, because all other terms cancel. So we'll have

$$\sum_{j = 0}^k (x_{j+1} - x_j) = x_{k+1} - x_0 = \beta^{k+1} - h^{k+1}.$$

Finally, since $\beta \neq h$ (so $\beta - h \neq 0$), we conclude

$$\sum_{j = 0}^k \beta^k h^{j-k} = \frac{\beta^{k+1} - h^{k+1}}{\beta - h}.$$

 

[Euge]

The one where beta goes to infinity!

What do you mean by this?

 

[Manfred1999]

Dear Euge,

Thank you very much for your very detailed answer. I eventually could answer the question myself. However, your explanations were still really useful. Much appreciated.

Manfred

The expressions $j^{k-j}$ should be $h^{k-j}$. To prove the corrected formula, perform synthetic division. Alternatively, multiply the sum $\sum\limits_{j = 0}^n \beta^j h^{k-j}$ by $\beta - h$; this gives

$$(\beta - h)\sum_{j = 0}^k \beta^j h^{k-j} = \sum_{j = 0}^k (\beta - h)\beta^j h^{k-j} = \sum_{j = 0}^k [\beta^{j+1}h^{k-j} - \beta^j h^{k-(j-1)}] = \sum_{j = 0}^k (x_{j+1} - x_j),$$

where $x_j = \beta^{j}h^{k-(j-1)}$. The sum you see at end is known as a telescoping sum. When you expand the sum, the only term that "survive" are $x_{n+1}$ and $x_0$, because all other terms cancel. So we'll have

$$\sum_{j = 0}^k (x_{j+1} - x_j) = x_{k+1} - x_0 = \beta^{k+1} - h^{k+1}.$$

Finally, since $\beta \neq h$ (so $\beta - h \neq 0$), we conclude

$$\sum_{j = 0}^k \beta^k h^{j-k} = \frac{\beta^{k+1} - h^{k+1}}{\beta - h}.$$