Question

Is the 50^th partial sum S₅₀ of the alternating series \(\sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^{n - 1}}}}{n}} \) an overestimate or an underestimate of the total sum? Explain.

Short answer

S₅₀ is an underestimate.

Step-by-step solution

Considering Partial Sum:

Consider the 50^th partial sum S₅₀ of the following series:

\(\sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^{n - 1}}}}{n}} \)

Determine S₅₀ is an underestimate or overestimate to the sum of the entire series,

\(S = \sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^{n - 1}}}}{n}} \)

First, note that this is clearly a decreasing alternating series, whose terms approach to zero.

The expression (-1)^n-1 alternates, and the factor \(\frac{1}{n}\) decreases and approaches to zero.

A general pattern for the partial sums of decreasing alternating series, the first partial sum starts positive, the second partial sum goes down, the third partial sum rises back not as high as the first, the fourth partial sum falls but not as high as the second and so on.

So, the sequence of odd terms is falling down to the limit, while the sequence of even terms is rising up to the limit.

For example, the n^th term of the sequence is \({a_n} = \frac{{{{( - 1)}^{n - 1}}}}{n}\)

\({\rm{Let }}n = 1{\rm{ then }}{S_1} = {a_1}\)

so

\({{\rm{a}}_1} = \frac{{{{( - 1)}^{1 - 1}}}}{1} = 1\)

\({\rm{Let }}n = 2{\rm{ , then }}{{\rm{S}}_2} = {S_1} + {a_2}\)

\({\rm{So }}{S_1} + {a_2} = 1 + \frac{{{{( - 1)}^{2 - 1}}}}{2} = \frac{1}{2}\)

\({S_2} = \frac{1}{2}\)

Calculating S3 and S4:

\(\begin{aligned}{S_3} &= {S_2} + {a_3} = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}{\rm{ rising back up but not as high as the first}}\\{S_4} &= {S_3} + {a_4} = \frac{5}{6} - \frac{1}{4} = \frac{7}{{12}}{\rm{ falling, but not as low as the second }}\end{aligned}\)

The expression (-1)^n-1 is positive on odd terms and negative on even terms. The number 50 is even, so the last term added in S₅₀ is negative and by the above pattern, expect that S₅₀ is an underestimate.

The proof is given below.

Note that a₅₁ being an odd term is positive and a₅₂ is negative.

Then \({a_{51}} + {a_{52}} = \left| {{a_{51}}} \right| - \left| {{a_{52}}} \right|\) since, as noted a₅₁ is positive and a₅₂is negative and

\(\left| {{a_{51}}} \right| - \left| {{a_{52}}} \right| \ge 0{\rm{ because }}\left| {{a_{51}}} \right| \ge \left| {{a_{52}}} \right|\)

\({\rm{So, }}{a_{51}} + {a_{52}} \ge 0\)

Conclusion from the Series:

Clearly, there is nothing special about a₅₁ and a₅₂ this holds for every pair of an odd term a_2n+1 and the succeeding even term a_2n+2 because of the combination of the alternating and decreasing property of the series.

Now, we notice the expression

\(\begin{aligned}\sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^{n - 1}}}}{n}} &= {S_{50}} + {a_{51}} + {a_{52}} + {a_{53}} + {a_{54}} + .... + {a_{2n - 1}} + {a_{2n - 2}} + ...\\{\rm{ }} &= {S_{50}} + ({a_{51}} + {a_{52}}) + ({a_{53}} + {a_{54}}) + .... + ({a_{2n + 1}} + {a_{2n + 2}}) + ...\end{aligned}\)

So we noted, each of the pairs in parenthesis is an odd term and the succeeding even term S₀ is positive.

That means, the sum of the entire series is S₅₀ plus a bunch of positive terms.

Therefore, S₅₀ is an underestimate.