∞ ∑ \infty\sum (ii)

Calculus Level 2

G i v e n t h e i n f i n i t e s u m : \small\mathrm{Given\ the\ infinite\ sum:} S = 1 + 2 2 + 1 2 3 + 1 + 3 2 + 1 3 3 + 1 + 4 2 + 1 4 3 + 1 ⋯ \small{S= 1+\frac{2^2+1}{2^3+1}+\frac{3^2+1}{3^3+1}+\frac{4^2+1}{4^3+1}\cdots}

W h a t c a n w e s a y a b o u t t h i s s u m ? \small\mathrm{What\ can\ we\ say\ about\ this\ sum?}

R u l e : S o l v e i t w i t h o u t u s i n g t h e c o m p a r i s o n t e s t \small\mathrm{\mathbf{Rule}: Solve\ it\ without\ using\ the\ comparison\ test}

Y e s ( W r i t e y o u r s o l u t i o n ) \small\mathrm{Yes\ (Write\ your\ solution)} M a y b e ? ( R e a l l y ? ) \small\mathrm{Maybe?\ (Really?)} W h a t ? ( S e e h o w i t s o l v e s ) \small\mathrm{What?\ (See\ how\ it\ solves)} N o ? ( T r y i t ! ) \small\mathrm{No?\ (Try\ it!)}

Victor Moreno Diaz by Victor Moreno Diaz , 25, Spain

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1 solution

Richard Polak
Jan 18, 2015

Even without using comparison test, we can grasp that the series behaves like the harmonic series. Thus, we can try Cauchy's condensation test (which is used to prove that the harmonic series diverges).

S = ∑ 1 ∞ n 2 + 1 n 3 + 1 d i v e r g e s i f ∑ 1 ∞ 2 n ( 2 n ) 2 + 1 ( 2 n ) 3 + 1 d i v e r g e s S=\sum_{1}^{\infty}\frac{n^2+1}{n^3+1} \ \ diverges \ if \ \ \sum_{1}^{\infty} 2^n\frac{(2^n)^2+1}{(2^n)^3+1} \ \ diverges

The general term of this series is bigger than 1 for all values of n: 2 n ( 2 n ) 2 + 1 ( 2 n ) 3 + 1 = 2 3 n + 2 n 2 3 n + 1 > 1 2^n\frac{(2^n)^2+1}{(2^n)^3+1} = \frac{2^{3n}+2^n}{2^{3n}+1} > 1

And 1+1+1+1+1... diverges so S diverges.

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