Verified answer

The shortest way is to use the Cauchy's Integral Test:

http://en.wikipedia.org/wiki/Maclaurin%E2%80%93Cau...

(Your example is in 'Applications' Section in the Article).

Somewhat longer is to prove it directly:

(ii) The divergence for p < 1 follows from the divergence for p = 1 - remove the 1st and the 2nd terms from the harmonic series and combine the rest like this:

(1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + (1/9 + . . + 1/16) + (1/17 + . . + 1/32) + . .

i.e. 2 terms, then 4 terms, then 8, etc. (in each pair of parentheses the number of addends is a power of 2). Now the sum in each pair of parentheses is greater than 2/4, 4/8, 8/16, etc., i.e. greater than 1/2, what implies that the partial sums of the harmonic series is unbounded and tends to infinity.

(i) Let p > 1 and p = 1 + s, s > 0, then obviously

(*) 1/(n+1)^p + 1/(n+2)^p + . . + 1/(2n)^p < n * (1/n^p) = 1/n^s

Combining similarly

(1/3^p + 1/4^p) + (1/5^p + 1/6^p + 1/7^p + 1/8^p) + (1/9^p + . . + 1/16^p) + . .

and applying (*) follows that the sum in each pair of parentheses is less than the corresponding term of the convergent geometric progression 1/2^s, 1/4^s, 1/8^s, .. etc., hence every partial sum is less than the sum of the aforementioned progression plus 1 + 1/2^p, i.e. the p-series converges (and its sum is the Riemann Zeta function ζ(p):

http://en.wikipedia.org/wiki/Zeta_function

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