How do you test for convergence for #1/((2n+1)!) #?
↳Redirected from
"What polyatomic ion has a positive charge?"
In the case you meant "test the convergence of the series : #sum_(n=1)^(oo)1/((2n+1)!)#"
the Answer is : it #color(blue)"converges"#
To find out, we can use the ratio test.
That is, if #"U"_"n"# is the #n^"th"# term of this series
Then if, we show that #lim_(nrarr+oo)abs("U"_("n"+1)/"U"_n)<1#
it means that the series converges
On the other if #lim_(nrarr+oo)abs(("U"_("n"+1))/"U"_n)>1#
it means that the series diverges
In our case
#"U"_n=1/((2n+1)!)#
#" "# and
#"U"_("n"+1)=1/([2(n+1)+1]!)=1/([2n+3]!)#
Hence, #"U"_("n"+1)/"U"_n=1/((2n+3)!)÷1/((2n+1)!)=((2n+1)!)/((2n+3)!)#
#"Notice that" :#
#(2n+3)! =(2n+3)xx(2n+2)xx(2n+1)!#
Just like : #10! =10xx9xx8!#
We subtract #1# each time to get the next
So we have,
#"U"_("n"+1)/"U"_n=((2n+1)!)/((2n+3)(2n+2)(2n+1)!)=1/((2n+3)(2n+2))#
Next we test,
#lim_(nrarr+oo)abs("U"_("n"+1)/"U"_n)#
#=lim_(nrarr+oo)abs(1/((2n+3)(2n+2)))=lim_(nrarr+oo)1/((4n^2+10n+6))=1/(+oo)=0" "# and #0# is less than #1#
Hence, it's quite safe to conclude that the series #color(blue)"converges" !#
