Determine whether the following series converges or diverges
∞
∑ [(-3/∏)^n + (e/9)^(n+1)]
n=1
If it converges, find its sum.
I don't know how to go about solving this one, can someone explain the steps that need to be taken?
Determine whether the following series converges or diverges
∞
∑ [(-3/∏)^n + (e/9)^(n+1)]
n=1
If it converges, find its sum.
I don't know how to go about solving this one, can someone explain the steps that need to be taken?
Distribute the sum over both terms, and if the series converges, both of those terms should converge.
If ∑a_{n }and ∑b_{n} are convergent series, then so are the series ∑ca_{n} (where c is a constant)
Then ∑(a_{n} + b_{n}) = ∑a_{n + }∑b_{n}
Correct they are both geometric series. Remember the sum of such a series ∑ ar^{n-1} = a/(1-r) provided that |r|<1
You need to get them into a form where you can easily determine the values of a and r. If |r| > 1 then the series diverges.
Lets do the using partial sums.
If $\displaystyle S_n = \sum\limits_{k = 1}^n {\left[ {\left( {\frac{{ - 3}}{\pi }} \right)^k + \left( {\frac{e}{9}} \right)^k } \right]} $ would you agree that $\displaystyle S_n=U_n+V_n$ where $\displaystyle U_n = \sum\limits_{k = 1}^n {\left[ {\left( {\frac{{ - 3}}{\pi }} \right)^k } \right]} $ and $\displaystyle V_n = \sum\limits_{k = 1}^n { {\left( {\frac{e}{9}} \right)^k } } $
Does each of $\displaystyle U_n~\&~V_n$ converge? If so, the sum of their limits is the sum of the original series.
Same idea as the prior, instead of decreasing n by 1 we now want to decrease n by 2.
(e/9)^(n-1) = (e/9)^2(e/9)^(n-1)
Not that if you add the exponents together you get 2 + (n - 1) = n + 1
This is always your goal with a geometric series. Always check that the absolute value of r is less than 1, because if r is greater than 1 the series diverges.