This is a comment at the comments of Gerhard "Still Computing Oh So Slowly" Paseman's answer.
I used the function fharm(n,eps) = frac(h(n))*n^(1+eps) to find the local minima for a fixed eps and consecutive arguments n. The n-1 of my list should agree with the n in the comments. (The harmonic numbers can be computed in Pari/GP using h(n) = psi(1+n) + Euler )
This is what I got using Pari/GP and 800 digits internal precision in a couple of seconds:
n | fharm(n,0.1)
--------------------+----------------
1 0.0
11 0.277901138881
31 1.19066007386
83 0.267050685436
227 1.70522868276
616 1.49190552481
1674 1.70759404228
4550 2.19789655729
12367 1.36471771489
33617 1.68810700160
91380 0.873923093872
248397 1.05774925083
675214 3.52030034776
1835421 2.93703145896
4989191 2.22651939403
13562027 1.04059409912
36865412 0.783097603793
100210581 6.15552338765
272400600 3.07296513521
740461601 2.29064626275
2012783315 2.37389093860
5471312310 1.01950811681
14872568831 3.51618209965
40427833596 2.82622840722
109894245429 10.6250101261
298723530401 7.64454034138
812014744422 6.96806150688
2207284924203 3.81951285918
6000022499693 2.46795156044
16309752131262 12.4771965474
44334502845080 1.85826500223
120513673457548 14.1650514186
327590128640500 21.4949794587
890482293866031 0.75550636057
2420581837980561 21.3699821691
6579823624480555 3.02859881920
17885814992891026 3.21976135398
48618685882356024 1.64093320527
132159290357566703 13.7707827691
359246197441016284 14.6487253696