How do you find the n-th partial sum of an infinite series?

Answer 1

To find the n-th partial sum of an infinite series, you add up the first n terms of the series. Mathematically, this can be represented as:

[ S_n = a_1 + a_2 + a_3 + \ldots + a_n ]

Where ( S_n ) is the n-th partial sum and ( a_1, a_2, a_3, \ldots ) are the terms of the series.

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Answer 2
The nth partial sum #S_n# of a series #sum_{k=1}^infty a_k# is #S_n=sum_{k=1}^n=a_1+a_2+a_3+cdots+a_n#
Let us find #S_n# for the telescoping series
#sum_{k=1}^infty(1/k-1/{k+1})#.

The partial sum is

#S_n=(1/1-1/2)+(1/2-1/3)+cdots+(1/n-1/{n+1})#

by regrouping,

#=1+(-1/2+1/2)+cdots+(1/n-1/n)-1/{n+1}#
#=1+0+cdots+0-1/{n+1}#
#=1-1/{n+1}=n/{n+1}#
Hence, #S_n=n/{n+1}#.
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Answer from HIX Tutor

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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