How do you find the nth partial sum, determine whether the series converges and find the sum when it exists given #1-2+3-4+...+n(-1)^(n-1)#?

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Answer 1 · 2018-01-04T02:20:33

#sum_(n=1)^oo n(-1)^(n-1)#

is indeterminate.

The series does not converge as it does not satisfy the necessary condition for convergence:

#lim_(n->oo) a_n = 0#

We can evaluate the partial sums noting that for #n# even we can group the terms as:

#s_(2n) = (1-2)+(3-4)+...(2n-1-2n) =underbrace(-1-1+...-1)_(n " times") =-n#

while for #n# odd:

#s_(2n+1) = 1 + (-2 +3) + (-4+5)+...+(-2n+2n+1) = underbrace(1+1+1+...+1)_(n+1 " times") = n+1#

Hence the sums oscillate and are unbounded in absolute value so that:

#lim_(n->oo) s_n#

does not exist.