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

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Answer 1 · 2017-02-01T00:57:13

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 $n->oo$ .

$sum_(k=0)^oo(1/(k!)-1/((k+1)!)) = lim_(n->oo)sum_(k=0)^n(1/(k!)-1/((k+1)!))$

$=lim_(n->oo)(1-1/((n+1)!))$

$=1-1/oo$

$=1-0$

$=1$

Answer 2 · 2017-02-01T01:05:44

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,

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

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)!))+...(1-1/2)+(1/2-1/6)+(1/6-1/24)+...+(1/(n!)-1/((n+1)!))+...?