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Worked solution

Find the first harmonic partial sum above four

Find the least positive integer nn for which ∑r=1n1/r>4\sum_{r=1}^{n}1/r>4.

CJC · Sequences and Series · Example 15

Answer

n=31n=31.

Compare your reasoning

Tabulate the partial sum as a function of its upper limit

Y1(X)=∑R=1X1RY_1(X)=\sum_{R=1}^{X}\frac1R

The calculator screen enters the sum as Y1 with upper limit X and a separate summation counter R. Evaluate at positive integer X values.

Show the adjacent values around the threshold

n∑r=1n1/r (3 d.p.)comparison with 4303.995<314.027>324.058>\begin{array}{c|c|c}n&\sum_{r=1}^{n}1/r\text{ (3 d.p.)}&\text{comparison with }4\\\hline30&3.995&<\\31&4.027&>\\32&4.058&>\end{array}

The source callout requires this supporting table for the calculator method. The printed n=32 entry 4.059 is corrected here to 4.058 after rounding the unrounded sum; the conclusion is unchanged.

Explain why the first crossing is minimal

Hn+1−Hn=1n+1>0,H30<4<H31H_{n+1}-H_n=\frac1{n+1}>0,\qquad H_{30}<4<H_{31}

Each added reciprocal is positive, so partial sums strictly increase. Since H30<4<H31, no earlier positive index works. This monotonicity argument supplements the printed table.

How did this question go?

Common mistakes

The unknown n is the number of included terms, not the summation counter. A table needs neighbouring values and increasing partial sums to justify the least index.

Solution notes

The original table prints 3.995, 4.027 and 4.059. The last value is a source rounding error: H32≈4.058495195437 rounds to 4.058 at 3 decimal places. The screenshot displays 4.0585; retain unrounded values when rounding again.