Prove the divergence of $\sum_{k=0}^{\infty}{\frac{1}{10k+1}}$

Question

Prove the divergence of $\sum_{k=0}^{\infty}{\frac{1}{10k+1}}$. By the ratio test:$$\frac{\frac{1}{10(k+1)+1}}{\frac{1}{10k}}=\frac{10k}{10(k+1)+1}=\frac{10k}{10k+11}=\frac{10k}{k(10+\frac{11}{k})}=\frac{10}{10+\frac{11}{k}}\overset{k\to \infty}\longrightarrow 1$$ thus it diverges for almost all $k\in \mathbb{N}$

Is that correct?

Answer 1

What you have is incorrect since there is no notion of convergence for almost all $k$ due to the ratio test. In fact, since your limit goes to $1$ as you correctly show, the test is inconclusive so the series could converge or diverge... the thing to recognize here is that the application of the ratio test didn't provide any extra information.

However, if we observe $$\sum_{k=0}^\infty\frac{1}{10k+1}=1+\sum_{k=1}^\infty\frac{1}{10k+1},$$ we need only study the convergence/divergence of $\sum_{k=1}^\infty(10k+1)^{-1}$. It is clear that $10k+1\leq 10k+k=11k$ for every $k\geq1$, so we have by taking reciprocals that $$\frac{1}{10k+1}\geq\frac{1}{11k}.$$ In particular, we may sum over all $k\geq1$ on each side of the inequality to obtain $$\sum_{k=1}^\infty\frac{1}{10k+1}\geq\sum_{k=1}^\infty\frac{1}{11k}=\frac{1}{11}\left(\sum_{k=1}^\infty\frac{1}{k}\right).$$Since this latter series is the harmonic series, it diverges to $+\infty$, implying that the sum on the left also diverges to $+\infty$; in particular, this applies to the original sum.

Answer 2

You can use the limit comparison test with b$_k$ = $\frac{1}{10k}$. Sine the limit of $\mid \frac{a_k}{b_k} \mid$ is finite, both series do the same thing, namely diverge.

Or you can use the integral test and show that the improper integral diverges.

Answer 3

Observe that \begin{align*} \sum_{k=0}^\infty \dfrac{1}{10k+1} &= 1 + \frac{1}{11} + \frac{1}{21} + \frac{1}{31}+ \frac{1}{41}+ \frac{1}{51}+ \frac{1}{61} + \cdots \\ &\geq \frac{1}{32} + \frac{1}{64} + \frac{1}{64} + \frac{1}{128}+ \frac{1}{128}+ \frac{1}{128}+ \frac{1}{128}+\cdots \\ &= \sum_{k=0}^\infty 2^{-5-\lfloor \log_2(k+1) \rfloor} \text{,} \end{align*} where $\lfloor x \rfloor$ is the floor function, returning the largest integer less than or equal to $x$.

If this last series diverges, then by the direct comparison test, so does the given series.

The terms of this last series are one copy of $1/32$, two copies of $1/64$, four copies of $1/128$, eight copies of $1/256$, and so on, having $2^n$ copies of $2^{-n-5}$ as $n$ increases through the integers, starting at $0$. Notice that, for each of these runs of identical terms, the sum of the terms of that run is always $2^k \cdot 2^{-k-5} = 1/32$. Also, all the terms are positive. Therefore, the sequence of partial sums of the latter series is strictly monotonically increasing and meets every positive multiple of $1/32$. In particular, the partial sum is $n/32$ when the first $n$ runs of terms are summed. Since the sequence of positive multiples of $1/32$ increases without bound, the sequence of partial sums of the last series increases without bound, hence diverges. Therefore, the given series diverges.

The easiest way to show $\dfrac{1}{10k+1} \geq 2^{-5-\lfloor \log_2(k+1) \rfloor}$ is to show that $$ 2^{-5-\lfloor \log_2(k+1) \rfloor} \leq 2^{-4-\log_2(k+1)} = \frac{1}{16k + 16} \text{,} $$ by thinking about what interval around $x$ the result of $\lfloor x \rfloor$ always resides in. Then showing $16k + 16 \geq 10k + 1$ for $k \geq 0$ is easy.