How do you find the nth partial sum, determine whether the series converges and find the sum when it exists given #(1-1/2)+(1/2-1/6)+(1/6-1/24)+...+(1/(n!)-1/((n+1)!))+...#?

Answer 1

To find the nth partial sum of the series ( (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{6}) + (\frac{1}{6} - \frac{1}{24}) + \ldots + (\frac{1}{n!} - \frac{1}{(n+1)!}) + \ldots ), you can observe that each term can be expressed as the difference between two consecutive terms in the sequence ( \frac{1}{n!} ). Thus, the nth partial sum can be written as:

[ S_n = 1 - \frac{1}{(n+1)!} ]

To determine whether the series converges, you can take the limit of the nth partial sum as ( n ) approaches infinity:

[ \lim_{n \to \infty} S_n = \lim_{n \to \infty} \left(1 - \frac{1}{(n+1)!}\right) = 1 - \lim_{n \to \infty} \frac{1}{(n+1)!} = 1 ]

Since the limit of the nth partial sum exists and is finite, the series converges.

To find the sum of the series when it exists, you use the fact that the limit of the nth partial sum approaches the sum of the series as ( n ) approaches infinity. Therefore, the sum of the series is:

[ \sum_{n=1}^{\infty} \left(1 - \frac{1}{(n+1)!}\right) = 1 ]

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Answer 2

The #n^"th"# partial term is given by

#sum_(k=1)^n(1/(k!)-1/((k+1)!)) = 1-1/((n+1)!)#

and the series converges to #1# as #n->oo#.

First, to find the #n^"th"# partial sum:
#sum_(k=1)^n(1/(k!)-1/((k+1)!)) = (1/(1!)-1/(2!))+(1/(2!)-1/(3!))+...+(1/(n!)-1/((n+1)!))#
#=1+(-1/(2!)+1/(2!))+(-1/(3!)+1/(3!))+...+(-1/(n!)+1/(n!))-1/((n+1)!)#
#=1-1/((n+1)!)#

A series like this, in which all but a finite number of terms cancel out, is called a telescoping series.

With our closed form for the #n^"th"# partial sum, we can now show that the series converges to #1# as #n->oo#.
#sum_(k=0)^oo(1/(k!)-1/((k+1)!)) = lim_(n->oo)sum_(k=0)^n(1/(k!)-1/((k+1)!))#
#=lim_(n->oo)(1-1/((n+1)!))#
#=1-1/oo#
#=1-0#
#=1#
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Answer 3

The series converges to unity.

The nth partial sum, #S_n#, is given by:
# S_n = sum_(r=1)^(n) (1/(r!)-1/((r+1)!)) # # " "= (1-1/2) + # # " " (1/2-1/6)+# # " "(1/6-1/24)+...+# # " "(1/((n-1)!)-1/(n!)) + # # " "(1/(n!)-1/((n+1)!))#

Note that almost all the terms cancel, leaving:

# S_n = 1 -1/((n+1)!) #
As #n rarr oo => (n+1)! rarr oo => 1/((n+1)!) rarr 0 #

And so,

# lim_(n rarr oo) S_n = lim_(n rarr oo) (1 -1/((n+1)!)) # # " " = lim_(n rarr oo) (1) - lim_(n rarr oo)(1/((n+1)!)) # # " " = 1 - 0 # # " " = 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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