Question

approximate the sum of the series correct to four decimal places

Answer 1

??

Answer 2

what r choices?

Answer 3

\[\sum_{n=1}^{?\infty}\frac{ (-1)^{n}n }{ 11^{n}}\]

Answer 4

This is an alternating series

Answer 5

ok but how do I do this?

Answer 6

There's a theorem to this, \[\sum (-1)^n b_n \] is a convergent series then we can use the following \[|R_n|=|s-s_n| \le b_{n+1}\]\[b_n = \frac{ n }{ 11^n }\]

Answer 7

You could just plug in numbers

Answer 8

plug numbers until when?

Answer 9

Your n starts at 1, so you plug them in until you get to 4 decimal places as requested

Answer 10

Recall that for \(|x|<1\),
\[\sum_{n\ge1}x^{n-1}=\frac{1}{1-x}\]Let \(f(x)\) be this sum.

You have
\[f(x)=\frac{1}{x}\sum_{n\ge1}x^n\]Differentiate once and some manipulating gives
\[\begin{align*}f'(x)&=-\frac{1}{x^2}\color{red}{\sum_{n\ge1}nx^{n-1}}+\frac{1}{x}\sum_{n\ge1}x^n\\[1ex]
-\frac{x^2}{(1-x)^2}&=\color{red}{\sum_{n\ge1}nx^{n-1}}-x\sum_{n\ge1}x^n\\[1ex]
-\frac{x^2}{(1-x)^2}&=\color{red}{\sum_{n\ge1}nx^{n-1}}-x^2\sum_{n\ge1}x^{n-1}\\[1ex]
-\frac{x^2}{(1-x)^2}&=\color{red}{\sum_{n\ge1}nx^{n-1}}-\frac{x^2}{1-x}\\[1ex]
\frac{x^2(1-x)-x^2}{(1-x)^2}&=\color{red}{\sum_{n\ge1}nx^{n-1}}\\[1ex]
-\frac{x^3}{(1-x)^2}&=\color{red}{\sum_{n\ge1}nx^{n-1}}\\[1ex]
\end{align*}\]
Plug in \(x=-\dfrac{1}{11}\) and you can find the exact value of the series.

Answer 11

\[-\frac{ 1 }{ 11 } + \frac{ 2 }{ 121 }-...\]

Answer 12

it says "correct to four decimal places" and there is the 1/11^n. so all u have to do is to evaluate for n=1, 2 and 3....may be 4 just to be on the safe side.

Answer 13

ok the answer is -0.0764 how??

Answer 14

No, I mean you have to keep doing the pattern up until you get 4 decimal places, 0.0000some integers in your pattern

Answer 15

i got the answer wrong so it showed me the correct answer and it was -0.0764 but how do i plug that into my calculator to figure it out?

Answer 16

i think -0.0764 is the right ans :)

Answer 17

but ok how do i plug that into my calculator to get that answer?

Answer 18

But, you just have to plug in integers from your series

Answer 19

what do i plug into my calculator to get the answer -0.0764

Answer 20

-1/11+2/121 -3/11^3 +4/11^4

try it...it works - i just did.

Answer 21

oh sorry i didnt understand. but awesome I got the answer thanks guys sorry for being so difficult

Answer 22

Yw