How to find the sum of the series when a_n = (5^3x)*(7^(1-x))?

[biogirl]

How do I compute the sum of the series when a_n = (5^3n)(7^(1-n)) from n = 1 to infinity? I normally don't have much difficulty calculating sums, but I'm altogether lost with this one.

 

[stapel]

How do I compute the sum of the series when a_n = (5^3n)(7^(1-n)) from n = 1 to infinity? I normally don't have much difficulty calculating sums, but I'm altogether lost with this one.

What you've posted means this:

. . . . .\(\displaystyle a_n\, =\, \left(5^3 n\right)\left(7^{1-n}\right)\)

Is that what you meant? Or did you mean this second form?

. . . . .\(\displaystyle a_n\, =\, \left(5^{3 n}\right)\left(7^{1-n}\right)\)

If you mean the second, then there's a nice solution method.

Also, your subject line uses "x" and your post uses "n". Which is correct?

 

[biogirl]

I did mean the second form, yes, with 5^(3n). And n was the correct form, not x.

 

[Deleted member 4993]

How do I compute the sum of the series when a_n = (5^3n)(7^(1-n)) from n = 1 to infinity? I normally don't have much difficulty calculating sums, but I'm altogether lost with this one.

Please confirm that the second term is 7^1-n and NOT 7^1-3n

 

[HallsofIvy]

How do I compute the sum of the series when a_n = (5^3n)(7^(1-n)) from n = 1 to infinity? I normally don't have much difficulty calculating sums, but I'm altogether lost with this one.

a_n= (5^{3n})(7^{n+1})= (5^3)^n(7^{-n}(7)= 7(125/7)^n= is a geometric series of the form "\(\displaystyle \sum ar^n\)" with a= 7, r= 125/7.

It should be, in any case, clear that since 125/7 is greater than 1, this series will not converge.

 

[biogirl]

Ah thank you so much! It's obvious now.

 

[stapel]

Ah thank you so much! It's obvious now.

For readers who may not yet "see" the solution, here's a little hint:

. . . . .\(\displaystyle \displaystyle{\left(5^{3 n}\right)\left(7^{1-n}\right)\, =\, \left(5^3\right)^n\, \left(\frac{7}{7}\times \frac{1}{7^{n-1}}\right) }\)

. . . . .\(\displaystyle \displaystyle{=\, \left(5^3\right)^n\, \left(\frac{7}{7^n}\right)\, =\, \left(\frac{\left(5^3\right)^n}{7^n}\right)(7)}\)

...and so forth.