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Modifications of the Harmonic Series - Department of Mathematics

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Next, compare each with . The first nine terms <strong>of</strong> , are each less than<br />

a<br />

ai 1 a 2<br />

or equal to 1/ 10, so toge<strong>the</strong>r <strong>the</strong>y are less than 9/ 10.<br />

The next nine terms are each<br />

less than or equal to 1/ 20 and toge<strong>the</strong>r less than 9 / 20 = (9 /10) ⋅ (1/ 2) . Continue this<br />

process, and it is clear that a2 9/10 1 a < ⋅ . Similarly,<br />

2<br />

a3 81/100 a1 (9 /10) 1 a<br />

< ⋅ = ⋅ .<br />

1<br />

Fur<strong>the</strong>rmore, (9 /10) . Then we can estimate <strong>the</strong> modified series from above<br />

i−<br />

a < ⋅a<br />

with<br />

i<br />

1<br />

2 3<br />

⎛ ⎞<br />

9 ⎛ 9 ⎞ ⎛ 9 ⎞<br />

S < a1+ ⎜ + ⎜ ⎟ + ⎜ ⎟ + ... ⎟a1<br />

= 10a1 ≈27.18<br />

.<br />

⎜10 ⎝10 ⎠ ⎝10 ⎠ ⎟<br />

⎝ ⎠<br />

Estimating from Below: To estimate <strong>the</strong> series from below, we use a method<br />

similar to <strong>the</strong> one for estimating from above. Here we partition <strong>the</strong> series differently<br />

and estimate each finite sub-series from <strong>the</strong> o<strong>the</strong>r side. We shift <strong>the</strong> partitions that we<br />

used above by one term, so that<br />

1 1 1 1<br />

b1<br />

= + +⋅⋅⋅+ +<br />

2 3 8 10<br />

1 1 1 1<br />

b2<br />

= + +⋅⋅⋅+ +<br />

11 12 88 100 .<br />

1 1 1 1<br />

b3<br />

= + +⋅⋅⋅+ +<br />

101 102 888 1000<br />

<br />

We divide <strong>the</strong> terms <strong>of</strong> b again into groups <strong>of</strong> nine, and compare <strong>the</strong>m to <strong>the</strong><br />

2<br />

terms in b . The first group consists <strong>of</strong> 1/11+<br />

1/12 + ... + 1/18 + 1/ 20 . Each <strong>of</strong> <strong>the</strong>se<br />

1<br />

terms is greater than or equal to 1/ 20, and <strong>the</strong>y are toge<strong>the</strong>r greater than 9/ 20.<br />

Likewise, <strong>the</strong> next nine terms are greater than or equal to<br />

7<br />

9 / 30 = (9 /10) ⋅ (1/ 3) , and

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