KindiakMath

An Unnatural Logarithm

Problem 1. Prove that the sequence \{ \gamma_n \} defined by

\displaystyle \gamma_n := \sum_{k=0}^n \frac 1{k+1} - \ln(n+1).

converges to some real number \gamma, called the Euler-Mascheroni constant. Numerically, \gamma \approx 0.577.

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Solution. Write the sum as an integral as follows:

\begin{aligned} \gamma_n = \sum_{k=0}^n \frac 1{k+1} - \ln(n+1) &= \sum_{k=0}^n \frac 1{k+1} - \int_0^{n} \frac 1{x+1}\, \mathrm dx \\ &= \sum_{k=0}^n \int_{k}^{k+1} \frac 1{k+1}\, \mathrm dx - \sum_{k=0}^n \int_{k}^{k+1} \frac 1{x+1}\, \mathrm dx \\&= \sum_{k=0}^n \int_{k}^{k+1} \left( \frac 1{k+1} - \frac 1{x+1} \right) \, \mathrm dx. \end{aligned}

We observe that for any x \in [k, k+1],

\displaystyle 0 \leq \frac 1{k+1} - \frac 1{x+1} \leq \frac 1{k+1} - \frac 1{k+2}.

Since each summand is non-negative, \{\gamma_n\} is non-decreasing. Using the upper-bound,

\begin{aligned} 0 \leq \gamma_n &\leq \sum_{k=0}^n \left( \frac 1{k+1} - \frac 1{k+2} \right) = 1 -\frac 1{n+1} \leq 1. \end{aligned}

Hence, \{ \gamma_n \} is a non-decreasing sequence that is bounded above by 1. By the monotone convergence theorem, \gamma_n \to \gamma for some \gamma \in [0, 1].

Problem 2. Given that \{a_n\} is a non-negative \mathbb R-sequence that converges to 0, prove that the series \sum_{k=0}^\infty (-1)^k a_k converges. This result is called the alternating series test.

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Solution. For any n \in \mathbb N, define the n-th partial sum by

\displaystyle s_n := \sum_{k=0}^n (-1)^k a_k.

It is clear that s_0 = a_0 \geq 0 and s_1 = a_0 - a_1 \geq 0 since \{ a_n \} is decreasing.

For general m,

s_{2(m+1)} = s_{2m+2} = s_{2m} - \underbrace{ (a_{2m+1} - a_{2m+2}) }_{\geq 0} \leq s_{2m}.

Similarly, s_{2(m+1)+1} \geq s_{2m+1}. Combining these results, for any m \in \mathbb N,

\displaystyle s_1 \leq s_{2m+1} \leq s_{2m} \leq s_0.

Therefore, the sequence \{s_{2m}\} is decreasing and bounded below by s_1 \geq 0. By the monotone convergence theorem, s_{2m} \to s for some s \in \mathbb R. Similarly, s_{2m+1} \to s' for some s \in \mathbb R. Furthermore,

\displaystyle s' = \lim_{m \to \infty} s_{2m+1} = \lim_{m \to \infty} (s_{2m} + (-1)^{2m+1} a_{2m+1}) = s + 0 = s.

Finally, we observe that for any n, n = 2 i + j for some i \in \mathbb N and j \in \{0, 1\}. Hence,

\displaystyle s_{2i+1} \leq s_{2i + j} \leq s_{2i}.

Taking n \to \infty yields i = (n-j)/2 \to \infty, so that s_{2i} \to s and s_{2i+1} \to s. By the squeeze theorem, s_n = s_{2i+j} \to s. Therefore, the desired series s_\infty converges to s.

Problem 3. Evaluate the series 1 - 1/2 + 1/3 + 1/4 + \cdots.

(Click for Solution)

Solution. Write the series as

\displaystyle 1 - \frac 12 + \frac 13 - \frac 14 + \cdots = \sum_{k=0}^\infty (-1)^k \frac 1{k+1}.

Since the sequence \{1/(k+1)\} is non-negative and converges to 0, by the alternating series test, the series converges. For any natural number n,

\begin{aligned} \sum_{k=0}^{2n-1} (-1)^k \frac 1{k+1} &= \sum_{k=1}^{2n} \frac{1}{k} - \sum_{k=1}^{n} \frac{1}{k} \\ &= \sum_{k=0}^{2n-1} \frac{1}{k+1} - \sum_{k=0}^{n-1} \frac{1}{k+1} \\ &= (\gamma_{2n-1} + \ln(2n)) - (\gamma_{n-1} + \ln(n)) \\ &= \gamma_{2n-1} - \gamma_{n-1} + \ln(2).\end{aligned}

Taking n \to \infty,

\begin{aligned} \sum_{k=0}^{\infty} (-1)^k \frac 1{k+1} &= \lim_{n \to \infty} \sum_{k=0}^{2n-1} (-1)^k \frac 1{k+1} \\ &= \lim_{n \to \infty} (\gamma_{2n-1} - \gamma_{n-1} + \ln(2)) \\ &= \gamma - \gamma + \ln(2) \\ &= \ln(2). \end{aligned}

—Joel Kindiak, 12 Sept 25, 1732H

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joelkindiak

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