How do we find the sum of this series $\sum [a+(n-1)d] b r^{n-1}$?

Question

Let $a, b, d, r$ be real numbers such that $d \neq 0$ and $r \neq 0$. Let $$s_n \colon= [a+ (n-1)d] b r^{n-1}$$ for $n=1, 2, 3, \ldots$.

Then how do we find $$\sum_{n=1}^N s_n$$ for $N = 1, 2, 3, \ldots$?

Are there any values of $a, b, d, r$ for which the series $\sum s_n$ converges?

Are there any values of $a, b, d, r$ for which the series converges absolutely?

Answer 1

HINT You can start by writing it as $$(a-d)b\sum_{n=1}^Nr^{n-1}+db\sum_{n=1}^Nnr^{n-1}$$

the first sum is a geometric series and the second is the derivative of a geometric series

Answer 2

Based on David Quinn's hint:

Provided $r \neq 1$, we have the following calculation: $$\sum_{n=1}^N s_n = (a-d)b \sum_{n=1}^N r^{n-1} + db \sum_{n=1}^N nr^{n-1} = (a-d)b \frac{1-r^N}{1-r} + db \frac{ \mathrm{d} }{ \mathrm{d} r } \sum_{n=1}^N r^n = (a-d)b \frac{1-r^N}{1-r} + db \frac{ \mathrm{d} }{ \mathrm{d} r } \left( \frac{r- r^{ N+1 } }{1-r} \right) = (a-d)b \frac{ 1-r^N }{1-r} + db \frac{ [1-(N+1)r^N] (1-r) + r-r^{ N+1 } }{ (1-r)^2 } = (a-d)b \frac{ 1-r^N }{1-r} + db \frac{ 1 - (N+1)r^N + N r^{N+1} }{(1-r)^2} = \frac{ (a-d)b (1-r) (1-r^N) + db [ 1 - (N+1)r^N + N r^{N+1} ] }{ (1-r)^2} = \frac{ a( 1 - r - r^N + r^{N+1} ) - db [ 1 - r - r^N + r^{N+1} - ( 1 - (N+1)r^N + N r^{N+1} ) ] }{ (1-r)^2} = \frac{ a( 1 - r - r^N + r^{N+1} ) - db [ - r + N r^N - (N-1) r^{N+1} ] }{ (1-r)^2} = \frac{ a( 1 - r - r^N + r^{N+1} ) + dbr [ 1 - N r^{N-1} + (N-1) r^N ] }{ (1-r)^2} .$$

But how do we answer the remaining questions in my post?

Answer 3

you can break $$\sum_{n=1}^N s_n$$ as ab$\sum_{n=1}^N r^{n-1}$ +bd$\sum_{n=1}^N (n-1)r^{n-1}$

you can find the geometric sum $\sum r^{n-1}$....now you take k=$\sum_{n=1}^N (n-1)r^{n-1}$

see k-rk= $ r+r^2+r^3+r^4+..........+r^{N-2}$+(N-1)$r^N$=k(1-r)

now k=$\frac{r(1-r^{N-2})}{(1-r)^2}$ +$\frac{(N-1)r^N}{1-r}$

provided r is not 1

if r=1 then $s_n$ becomes ap series.