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MathHardAdv 2010 • Paper 1

Sum of Infinite Geometric Series with Modulus Terms

Let SkS_k, k=1,2,…,100k = 1, 2, \dots, 100, denote the sum of the infinite geometric series whose first term is k−1k!\frac{k - 1}{k!} and the common ratio is 1k\frac{1}{k}. Then the value of 1002100!+∑k=1100∣(k2−3k+1)Sk∣\frac{100^2}{100!} + \sum_{k=1}^{100} \left|(k^2 - 3k + 1)S_k\right| is

Official Numerical Answer3

Topics & Concepts

Sequences and SeriesGeometric ProgressionSum of Series

Step-by-Step Solution

To find the value of the given expression, we first analyze SkS_k, which is the sum of the infinite geometric series with first term a=k−1k!a = \frac{k - 1}{k!} and common ratio r=1kr = \frac{1}{k}.

Step 1: Compute SkS_k for each kk

For k=1k = 1:
The first term is a=1−11!=0a = \frac{1 - 1}{1!} = 0. Since all terms of the series are zero, the sum is:
S1=0S_1 = 0

For k≥2k \ge 2:
The common ratio is r=1kr = \frac{1}{k}, where ∣r∣<1|r| < 1. The sum of an infinite geometric series is given by Sk=a1−rS_k = \frac{a}{1 - r}:
Sk=k−1k!1−1k=k−1k!k−1k=kk!=1(k−1)!S_k = \frac{\frac{k - 1}{k!}}{1 - \frac{1}{k}} = \frac{\frac{k - 1}{k!}}{\frac{k - 1}{k}} = \frac{k}{k!} = \frac{1}{(k - 1)!}

Step 2: Evaluate ∣(k2−3k+1)Sk∣|(k^2 - 3k + 1)S_k| for each kk

  • For k=1k = 1:
    ∣(12−3(1)+1)S1∣=∣(−1)⋅0∣=0|(1^2 - 3(1) + 1)S_1| = |(-1) \cdot 0| = 0

  • For k=2k = 2:
    S2=1(2−1)!=1S_2 = \frac{1}{(2 - 1)!} = 1
    ∣(22−3(2)+1)S2∣=∣(4−6+1)⋅1∣=∣−1∣=1|(2^2 - 3(2) + 1)S_2| = |(4 - 6 + 1) \cdot 1| = |-1| = 1

  • For k≥3k \ge 3:
    The quadratic k2−3k+1=(k−1)(k−2)−1>0k^2 - 3k + 1 = (k - 1)(k - 2) - 1 > 0 for all k≥3k \ge 3. Thus, the modulus can be removed:
    ∣(k2−3k+1)Sk∣=k2−3k+1(k−1)!|(k^2 - 3k + 1)S_k| = \frac{k^2 - 3k + 1}{(k - 1)!}

We can rewrite the numerator k2−3k+1k^2 - 3k + 1 as (k−1)(k−2)−1(k - 1)(k - 2) - 1:
k2−3k+1(k−1)!=(k−1)(k−2)−1(k−1)!=(k−1)(k−2)(k−1)!−1(k−1)!=1(k−3)!−1(k−1)!\frac{k^2 - 3k + 1}{(k - 1)!} = \frac{(k - 1)(k - 2) - 1}{(k - 1)!} = \frac{(k - 1)(k - 2)}{(k - 1)!} - \frac{1}{(k - 1)!} = \frac{1}{(k - 3)!} - \frac{1}{(k - 1)!}

Step 3: Compute the sum ∑k=1100∣(k2−3k+1)Sk∣\sum_{k=1}^{100} |(k^2 - 3k + 1)S_k|

Splitting the sum for k=1,2k = 1, 2 and k≥3k \ge 3:
∑k=1100∣(k2−3k+1)Sk∣=0+1+∑k=3100(1(k−3)!−1(k−1)!)\sum_{k=1}^{100} |(k^2 - 3k + 1)S_k| = 0 + 1 + \sum_{k=3}^{100} \left( \frac{1}{(k - 3)!} - \frac{1}{(k - 1)!} \right)

The sum ∑k=3100(1(k−3)!−1(k−1)!)\sum_{k=3}^{100} \left( \frac{1}{(k - 3)!} - \frac{1}{(k - 1)!} \right) is telescoping:
∑k=3100(1(k−3)!−1(k−1)!)=(10!−12!)+(11!−13!)+(12!−14!)+⋯+(197!−199!)\sum_{k=3}^{100} \left( \frac{1}{(k - 3)!} - \frac{1}{(k - 1)!} \right) = \left( \frac{1}{0!} - \frac{1}{2!} \right) + \left( \frac{1}{1!} - \frac{1}{3!} \right) + \left( \frac{1}{2!} - \frac{1}{4!} \right) + \dots + \left( \frac{1}{97!} - \frac{1}{99!} \right)

Canceling all intermediate terms yields:
=10!+11!−198!−199!=1+1−(198!+199!)=2−(198!+199!)= \frac{1}{0!} + \frac{1}{1!} - \frac{1}{98!} - \frac{1}{99!} = 1 + 1 - \left( \frac{1}{98!} + \frac{1}{99!} \right) = 2 - \left( \frac{1}{98!} + \frac{1}{99!} \right)

Combining the tail terms:
198!+199!=99+199!=10099!=1002100!\frac{1}{98!} + \frac{1}{99!} = \frac{99 + 1}{99!} = \frac{100}{99!} = \frac{100^2}{100!}

Therefore, the total sum is:
∑k=1100∣(k2−3k+1)Sk∣=1+2−1002100!=3−1002100!\sum_{k=1}^{100} |(k^2 - 3k + 1)S_k| = 1 + 2 - \frac{100^2}{100!} = 3 - \frac{100^2}{100!}

Step 4: Calculate the required value

Substituting this back into the original expression:
1002100!+∑k=1100∣(k2−3k+1)Sk∣=1002100!+(3−1002100!)=3\frac{100^2}{100!} + \sum_{k=1}^{100} \left|(k^2 - 3k + 1)S_k\right| = \frac{100^2}{100!} + \left( 3 - \frac{100^2}{100!} \right) = 3

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