We studied several family of series in detail. We also learned quite a few tests that can tell us if a series converge or diverge indirectly without computing partial sum. We can also compare one series to another series whose convergence/divergence property is known.
Tianran Chen
Department of Mathematics
Auburn University at Montgomery
We learned quite a few tests that can tell us if a series is convergent or divergent indirectly without computing partial sum, e.g. integral tests.
But for many series, this can be difficult. E.g., \[ \sum_{k=1}^\infty \frac{1}{k^2 + 1} \]
We already know the series \[ \sum_{k=1}^\infty \frac{1}{k^2} \] is convergent. What can this tell us about the series above?
This test work by comparing two series term by term directly. For this to work nicely we have to restrict ourselves to series of nonnegative terms.
If $0 \le a_k \le c_k$ for all positive integer $k$, and $\sum_{k=1}^\infty c_k$ converges, then $\sum_{k=1}^\infty a_k$ also converges.
Since the terms are nonnegative, terms must shrink to 0 quickly enough for the series to converge; If each term is bounded by the corresponding term in a series whose terms are known to shrink to 0 sufficiently quickly, then we can conclude the original series also converge.
Since \[ \frac{1}{k^2 + 1} \quad < \quad \frac{1}{k^2} \]
we can conclude that\[ \sum_{k=1}^\infty \frac{1}{k^2 + 1} \text{ converges} \quad \Longleftarrow \quad \sum_{k=1}^\infty \frac{1}{k^2} \text{ converges} \]
Similarly, we can determine if a series is divergent by comparing it to a divergent series.
If $a_k \ge d_k \ge 0$ for all positive integer $k$, and $\sum_{k=1}^\infty d_k$ diverges, then $\sum_{k=1}^\infty a_k$ also diverges.
For example, since
We can conclude $\sum_{k=1}^\infty \frac{1}{k - \frac{1}{3}}$ is also divergent.
Use the direct comparison tests to determine the convergence properties of the following series:
The direct comparison test stated above is quite easy to use if we can find reference series that satisfy the hypothesis, which is not always easy. For example, we wouldn't be able to compare \[ \sum_{k=1}^\infty \frac{1}{k + 1} \quad\text{to}\quad \sum_{k=1}^\infty \frac{1}{k} \;\text{(harmonic series)} \] using direct comparison test, because $\frac{1}{k+1} < \frac{1}{k}$.
The limit comparison test replaces such direct comparisons with comparisons using limits, which can be much more forgiving.
Limit comparison test. Let $\sum_{k=1}^\infty a_k$ and $\sum_{k=1}^\infty b_k$ be two series with $a_k, b_k \ge 0$.
For example, applying this to $\sum_{k=1}^\infty \frac{1}{k}$ and $\sum_{k=1}^\infty \frac{1}{k + 1}$, since \[ \lim_{k \to \infty} \frac{ \frac{1}{k} }{ \frac{1}{k+1} } = \lim_{k \to \infty} \frac{ k + 1 }{ k } = \lim_{k \to \infty} 1 + \frac{ 1 }{ k } = 1, \] the two series either both converge or both diverge. But we already the harmonic series $\sum_{k=1}^\infty \frac{1}{k}$ diverges, thus the series $\sum_{k=1}^\infty \frac{1}{k + 1}$ also diverges.
Using the limit comparison test to determine the convergence properties of the following series. \begin{align*} &\sum_{k=1}^\infty \frac{2}{3k + 5} \\ &\sum_{k=1}^\infty \frac{3k^2 - 4k + 2}{2k^5 + 4k^3 + 1} \\ &\sum_{k=1}^\infty \frac{\sqrt{k}}{3k + 2} \end{align*}