How do you find the sum of #(1+1)+(1/3+1/5)+(1/9+1/25)+...+(3^-n+5^-n)+...#?
To find the sum of the series (1+1)+(1/3+1/5)+(1/9+1/25)+...+(3^-n+5^-n)+..., we can notice that each pair of terms within parentheses forms a geometric series. Specifically, (1+1) is equivalent to 2, (1/3+1/5) can be written as (1/3)(1+5/3), (1/9+1/25) can be written as (1/9)(1+25/9), and so on. In general, for the nth pair, it's (3^(-n))*(1+5^(-n)).
Each pair follows the geometric series formula: sum = a * (1 - r^n) / (1 - r), where 'a' is the first term, 'r' is the common ratio, and 'n' is the number of terms.
For the nth pair:
- First term (a) = 3^(-n)
- Common ratio (r) = 5^(-n)
Using the formula for the sum of a geometric series, we get the sum of each pair as: sum = (3^(-n)) * (1 - (5^(-n))^2) / (1 - 5^(-n))
Then, summing up all these individual sums from n=1 to infinity gives the total sum of the series.
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We have:
Both these series are geometric series so the sum is:
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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