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Worked solution

Find the first geometric partial sum above 2019

Let SnS_n be the sum of the first nn terms of 1+2+4+8+16+⋯1+2+4+8+16+\cdots. Find the least positive integer nn such that Sn>2019S_n>2019.

CJC · Sequences and Series · Example 24

Answer

n=11n=11.

Compare your reasoning

Form the partial-sum expression

Sn=20+21+22+⋯+2n−1=(20)(2n−1)2−1=2n−1.\begin{aligned}S_n&=2^0+2^1+2^2+\cdots+2^{n-1}\\&=\frac{(2^0)(2^n-1)}{2-1}\\&=2^n-1.\end{aligned}

Retain the source powers and finite-sum substitution.

Solve the strict inequality

Sn>2019,2n−1>2019,2n>2020,n>ln⁡2020ln⁡2≈10.980.\begin{aligned}S_n&>2019,\\2^n-1&>2019,\\2^n&>2020,\\n&>\frac{\ln2020}{\ln2}\approx10.980.\end{aligned}

The source displays the logarithmic cutoff approximately as 10.980. Keep its exact value before choosing an integer.

Choose the least whole number of terms

S10=1023<2019<2047=S11\begin{aligned}&S_{10}\\ & =1023<2019<2047\\ & =S_{11}\end{aligned}

The printed least index is 11. These adjacent integer checks independently confirm the strict crossing.

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