What is the partial sum of the series?

[Feynmanfan]

Hello everybody!

I'm having some trouble with series. My calculus teacher asked us to find the partial sum of

Sigma from 1 to n [n^-(1 + 1/n)]

It is obvious that the series diverges when trying to find the infinite sum. However, is it possible to find an expression dependant of n of the partial sum? I don't know where to start from

 

[Ebolamonk3y]

1/(n*n^1/n) ... how does that go on the tex thing? I couldn't get it to work...

 

[M98Ranger]

.

The partial sum of a series is the sum of a certain number of terms in the series, rather than the sum of all infinite terms. In this case, the partial sum of the series would be the sum of the first n terms, where n is a positive integer. To find the partial sum of this specific series, you can use the formula for the sum of a geometric series, which is Sn = a(1-r^n)/(1-r), where a is the first term and r is the common ratio. In this case, a = 1 and r = 1/n. So the partial sum would be Sn = 1(1-(1/n)^n)/(1-(1/n)). Simplifying this expression would give you the partial sum as n approaches infinity. However, since this series diverges, the partial sum would also approach infinity as n approaches infinity. Therefore, there is no finite expression for the partial sum of this series.