This series can be rewritten, after some algebraic changes, as geometric series:

Use the rules of exponents to rewrite `4^(n+1) = 4^n * 4`

and `e^(2n) = (e^2)^n`

Then,

`sum_(n=1)^oo ((4^(n+1))/e^(2n)) = sum_(n=1)^oo 4* (4^n)/(e^2)^n =`

`=4sum_(n=1)^oo (4/e^2)^n`

Now it is apparent that this is geometric series with the common ratio `r =...

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This series can be rewritten, after some algebraic changes, as geometric series:

Use the rules of exponents to rewrite `4^(n+1) = 4^n * 4`

and `e^(2n) = (e^2)^n`

Then,

`sum_(n=1)^oo ((4^(n+1))/e^(2n)) = sum_(n=1)^oo 4* (4^n)/(e^2)^n =`

`=4sum_(n=1)^oo (4/e^2)^n`

Now it is apparent that this is geometric series with the common ratio `r = 4/e^2`

` `which approximately equals `r = 4/9 `

So this geometric series converges because its common ratio is less than 1

and it sum can be found using the formula

`sum_(n=1)^oo a_n = a_1/(1 - r)`

Thus, `sum_(n=1) ^ oo (4/e^2)^n = (4/e^2)/(1 - 4/e^2) =`

`=4/(e^2 - 4)`

And the original series `sum_(n=1)^oo (4^(n+1)/(e^(2n))) = 4sum_(n=1)^oo (4/e^2)^n = 4*4/(e^2 - 4) = `

`=16/(e^2 - 4)`

**The sum of the given series is `16/(e^2 - 4).`**