Determine whether this series converges: $\sum_{k=1}^{\infty} \frac{\sqrt{k}}{k^2+3}$
I am thinking of using the comparison test to prove that the series converges. I will compare this series to $\frac{1}{k\sqrt{k}}$. Is this correct? I'm not sure how it works. If the comparison test does not work, can you suggest other tests that can be applied here?
On
The method for rational functions and some other functions, to find the right $p$-series for direct or limit comparison is the same: First find a value of the index after which all terms are positive. Then, from the numerator and denominator factor out the leading power of the index, leaving something whose terms have constant (mostly zero) limits. Using your problem as an example... $$ \frac{\sqrt{k}}{k^2+3} $$ The leading (highest) power of $k$ in the numerator is easy: $k^{1/2}$. The leading power in the denominator is also easy: $k^2$. We factor those out of the numerator and denominator, respectively. \begin{align*} \frac{\sqrt{k}}{k^2+3} &= \frac{k^{1/2}}{k^2} \cdot \frac{1}{1+3/k^2} \\ &= \frac{1}{k^{3/2}} \cdot \frac{1}{1+3/k^2} \text{.} \end{align*} We have essentially factored the general summand is "the power of $k$ that is most nearly parallel to the summand" times "a correction multiplier that rapidly approaches (in this case) $1$".
Okay... We factored it. So how do we use it? We "build back up" to the general term you have. We start by examining the nearly constant part. $$ 1 < 1 + 3/k^2 \leq 1+3/1^2 = 4 $$ That may seem fast. Building up this inequality is good warmup for the build-up we have after, so let's work through this one. We start with the low end of the index. \begin{align*} k &\geq 1 \\ k^2 &\geq 1^2 = 1 & &\text{squaring positives is monotonic} \\ \frac{1}{k^2} &\leq \frac{1}{1^2} = 1 & &\text{reciprocals reverse inequalities} \\ \frac{3}{k^2} &\leq \frac{3}{1^2} = 3 \\ 1 + \frac{3}{k^2} &\leq 1 + \frac{3}{1^2} = 4 \\ \end{align*} To get "$1 < 1 + 3/k^2$" we just use that $3$ and $k^2$ are positive, so their ratio is positive so we have added something bigger than $0$ to $1$ yielding something bigger than $1$.
So, where are we? $$ \frac{1}{k^{3/2}} \cdot 1 < \frac{1}{k^{3/2}} \cdot \frac{1}{1+3/k^2} \leq \frac{1}{k^{3/2}} \cdot 4 \text{.} $$ This strongly suggests a comparison test with the converging $p$-series $\sum k^{-3/2}$. The direct comparison test will succeed if the general term of your series is less than the general term of the converging $p$-series. (Positive and smaller than converging is converging. Positive and bigger than diverging is diverging.) So let's build up (again). \begin{align*} 1 &< 1 + \frac{3}{k^2} \\ k^{3/2} \cdot 1 &< k^{3/2} \cdot \left( 1 + \frac{3}{k^2} \right) \\ \frac{1}{k^{3/2} \cdot 1} &> \frac{1}{k^{3/2} \cdot \left( 1 + \frac{3}{k^2} \right)} \\ \frac{1}{k^{3/2}} &> \frac{1}{k^{3/2}} \cdot \frac{1}{1 + \frac{3}{k^2}} \end{align*}
The left-hand side is the general term of the given series. The right-hand side is the general term of the converging $p$-series. By the direct comparison test, the given series converges.
Suppose the inequality had gone the other way. (The easiest way to do this is to make the general term of the given series be $\displaystyle \frac{\sqrt{k}}{k^2 - 3}$. Then the direct comparison test will give us no information. (Positive and bigger than a converging series could be anything.)) The limit comparison test, however, will tell us the same thing the factored form $$ \frac{1}{k^{3/2}} \cdot \frac{1}{1 - 3/k^2} $$ tells us: The $p$-series $\sum k^{-3/2}$ most closely parallels this series up to a nearly constant multiplier between $-2$ and $1$. (This is an example of the first step way back at the beginning of this post. "First find a value of the index after which all terms are positive." $k = 2$ is such an index. So we should frame our series as the mixed signs part, $\displaystyle \sum_{k=1}^1 \frac{\sqrt{k}}{k^2 - 3}$ and the positive part, $\displaystyle \sum_{k=2}^\infty \frac{\sqrt{k}}{k^2 - 3}$. The finite sum is just some number, so does no affect convergence. The positive part then is subject to the above factoring and comparison tests.)
You're on the right track. Specifically, note that for all $k \ge 1$, we have $$\frac{\sqrt{k}}{k^2 + 3} < \frac{\sqrt{k}}{k^2}$$ becaues the LHS denominator is strictly larger than the RHS denominator. Then since the RHS simplifies to $$\frac{\sqrt{k}}{k^2} = \frac{1}{k^{3/2}},$$ we see that $$\sum_{k=1}^\infty \frac{\sqrt{k}}{k^2 + 3} < \sum_{k=1}^\infty \frac{1}{k^{3/2}}.$$ Since the sum on the right is convergent, and the sum on the left is obviously positive (because all of its terms are positive), the sum on the left must converge.