Mathematics Stack Exchange
Mathematics Stack Exchange

Questions tagged [geometric-series]

For questions about or involving geometric series, a series where successive terms have a common ratio.

A geometric series is of the form

$$\sum_{n=0}^{\infty}ar^n.$$

If $|r| < 1$, then the series converges to $\frac{a}{1-r}$. If $|r| \geq 1$, the series diverges.

886 questions
-1
votes
2 answers

Proof Of sum of geometric series goes wrong!!!

I'm really struggling with the way this book proves the sum of geometric series. As I know the formula of sum of geometric series should be $$s_n= a\frac{1-r^n}{1-r}$$ But the book wrote it differently as $$s_n= a\frac{1-r^{n+1}}{1-r}$$ How is this…
Nothing
  • 1,708
-2
votes
2 answers

Series $z^{n!}$ and the geometric series

For the geometric series $$ \sum_{n=0}^\infty z^n =\lim_{n \rightarrow \infty} \frac{1}{1-z^n} $$ for $$ |z|<1 $$ How does the series $$ \sum_{n=0}^\infty z^{n!} $$ compare? $$ \sum_{n=0}^\infty z^{n!} = \sum_{n=0}^\infty z^{n(n-1)!} = 1 + z + z^2 +…
Toma
  • 161
-3
votes
1 answer

Can someone explain to me why I cant pull out the constant when I take the variance?

can someone explain why this is wrong: $\epsilon_t$ is WN~ $(0,\sigma^2)$ $X_t=\frac{\epsilon_t}{1-\phi}$ $V(X_t)=V(\frac{\epsilon_t}{1-\phi})=\frac{V(\epsilon_t)}{(1-\phi)^2}$ $V(X_t)=\frac{\sigma^2}{(1-\phi)^{2}}$ Why is this wrong? $\phi <1$ is…
mth
  • 1
-3
votes
1 answer

Geometric series, distance travelled when the ball hits the ground for the 5th time

a ball dropped on the surface takes a sequence of vertical bounces with each bounce the ball loses 15ft its preceding height. the ball is dropped from 25 feet. find the total distance travelled when the ball hits the ground for the 5th time
Daichi Lim
  • 1
-3
votes
1 answer

Arithmetic and Geometric Sequence (Progression)

The common difference of an arithmetic sequence is 1, and the common ratio of a geometric sequence is 3. A new sequence is formed by adding the corresponding terms of these two sequences. Suppose that the second and fourth terms of the new…
nicetomeetyou98
  • 13
-3
votes
1 answer

Write a geometric series if $r = 1/6$ and $n = 8$.

Write a geometric series if $r = 1/6$ and $n = 8$. I tried using the formula $a_n = a_1r^{n-1}$ and $S_n = a_1\cfrac{(1-r^n)}{1-r}$ but I don't know how to find the values for the terms like $a_1$.
Nafisa
  • 1
-4
votes
4 answers

Why can't this series be looked as a geometric series?

Why can't this series $$\sum_{n=1}^{\infty}\frac1{n^n}$$ be looked as a Geometric Series with $r=\frac1{n}$? I am looking for answers other than "$r$ is supposed to be a fixed real number from the definition of geometric series". Thank you.
DMH16
  • 1,517
1 2 3 4 5 6
7