# Prove $\sum_{i=1}^{30}\frac{1}{10+i} >\frac{13}{12}$

Prove $$\sum_{i=1}^{30}\frac{1}{10+i} >\frac{13}{12}$$

My effort:

\begin{aligned} & \frac{1}{11}>\frac{1}{42} \\ & \frac{1}{12}>\frac{1}{42} \\ & \frac{1}{13}>\frac{1}{42} \\ & \vdots \\ & \frac{1}{40}>\frac{1}{42}\\ & \Rightarrow \sum_{i=1}^{30}\frac{1}{10+i}>\frac{30}{42}=\frac{5}{7} \end{aligned}

I could not reach the bound.

• $$\frac{10}{20}+\frac{10}{30}+\frac{10}{40} = \frac{13}{12}$$ Commented Jul 12 at 1:00

You are estimating a little too roughly. Instead, you can divide your summands into three rough groups as @peterwhy suggests. Firstly, for $$i \leq 10$$ $$\frac{1}{10+i} \geq \frac{1}{20}$$ for $$10 \leq i \leq 20$$ applies: $$\frac{1}{10+i} \geq \frac{1}{30}$$ for $$20 \leq i \leq 30$$ applies: $$\frac{1}{10+i} \geq \frac{1}{40}.$$ If you now add these inequalities together, you get the desired estimate $$\sum_{i=1}^{30} \frac{1}{10+i} > 10 \cdot \frac{1}{20} + 10 \cdot \frac{1}{30} + 10 \cdot \frac{1}{40} = \frac{13}{12}$$

Let $$f(x)=\dfrac1{10+\left \lceil x \right\rceil}$$. We know $$\sum_{n=1}^{30}\frac1{10+n}=\int_0^{30}f(x) \text {d}x$$ For all $$x$$ in $$[0,30]$$, $$f(x)=\dfrac1{10+\left \lceil x \right\rceil}>\dfrac1{11+x}$$ so $$\int_0^{30}f(x)\text {d}x>\int_0^{30}\frac1{11+x}\text {d}x=\ln(41/11)>13/12$$

• If you don't mean the counting measure, your conversion to the integral is not correct. And if you mean the counting measure, you cannot evaluate the integral in this way. Commented Jul 12 at 1:19
• What do you mean by counting dimension? Commented Jul 12 at 1:21
• desmos.com/calculator/epovjyijkm is a link to the calculations. Commented Jul 12 at 1:22
• The sum is equal to the integral of the blue function, which is bounded below by the integral of the red function. Commented Jul 12 at 1:28
• @Noctis FYI, the integral is exact. Regarding a formal argument, note for each integer $i$ that $i-1 \lt x\le i$ has $f(x)=\frac{1}{10+\lceil x\rceil}=\frac{1}{10+i}$. Thus, $\int_{i-1}^{i}f(x)dx=\frac{1}{10+i}$ as $f(x)$ is constant, apart from the point at $x=i-1$, but a single point's value doesn't affect the integral, and the width of the rectangle is $1$. Summing these integrals for $i$ from $0$ to $30$, inclusive, gives what the OP has stated, i.e., $\sum_{n=1}^{30}\frac1{10+n}=\int_0^{30}f(x)dx$. Commented Jul 12 at 1:39

$${\displaystyle\sum_{i=1}^{30} \frac{1}{10+i}\\=\displaystyle\sum_{i=1}^{10} \frac{1}{10+i}+\displaystyle\sum_{i=11}^{20} \frac{1}{10+i}+\displaystyle\sum_{i=21}^{30} \frac{1}{10+i}\\>10\times \frac{1}{10+10}+10\times \frac{1}{10+20}+10\times \frac{1}{10+30}\\=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\\=\frac{13}{12}}$$

The most likely intended method is what's suggested in peterwhy's comment, plus is also implemented in Noctis's answer and ican's answer, i.e., grouping the values into $$3$$ sets of $$10$$ consecutive elements each, and then using that each subset sum is more than $$10$$ times its smallest member.

Another way to group the elements (which requires more effort here, although it can be a quite effective method in other problems) is the first with the last, the second with the second last, etc., up to the $$15$$th with the $$16$$th ones, to get

\begin{aligned} \sum_{i=1}^{30}\frac{1}{10 + i} & = \sum_{i=1}^{15}\left(\frac{1}{10 + i} + \frac{1}{10 + (31 - i)}\right) \\ & = \sum_{i=1}^{15}\left(\frac{1}{10 + i} + \frac{1}{41 - i}\right) \\ & = \sum_{i=1}^{15}\frac{(41 - i) + (10 + i)}{(10 + i)(41 - i)} \\ & = \sum_{i=1}^{15}\frac{51}{410 + 31i - i^2} \end{aligned}

Next, there's

\begin{aligned} 410 + 31i - i^2 & \lt -i^2 + 32i + 410 \\ & = -(i^2 - 32i) + 410 \\ & = -(i^2 - 32i + 256 - 256) + 410 \\ & = -(i - 16)^2 + 256 + 410 \\ & \le -1 + 666 \\ & = 665 \end{aligned}

Alternatively, we can get a slightly better upper bound by noting for $$1 \le i \le 14$$ that

\begin{aligned} & \frac{1}{410 + 31i - i^2} - \frac{1}{410 + 31(i + 1) - (i + 1)^2} \\ & = \frac{1}{410 + 31i - i^2} - \frac{1}{440 + 29i - i^2} \\ & = \frac{(440 + 29i - i^2) - (410 + 31i - i^2)}{(410 + 31i - i^2)(440 + 29i - i^2)} \\ & = \frac{30 - 2i}{(410 + 31i - i^2)(440 + 29i - i^2)} \\ & \gt 0 \end{aligned}

This means the terms are strictly decreasing so, for $$1 \le i \le 14$$ again,

$$\frac{1}{410 + 31i - i^2} \gt \frac{1}{410 + 31(15) - 15^2} = \frac{1}{650} \;\;\to\;\; \frac{51}{410 + 31i - i^2} \gt \frac{51}{650}$$

However, even using the somewhat worse earlier denominator upper bound of $$665$$ leading to a lower bound of $$\frac{15}{665}$$ in the first set of equations gives

\begin{aligned} \sum_{i=1}^{30}\frac{1}{10 + i} & \gt \frac{15(51)}{665} \\ & = \frac{3(51)}{133} \\ & \gt \frac{9(17)}{135} \\ & = \frac{17}{15} \end{aligned}

Finally, we have

$$\frac{17}{15} - \frac{13}{12} = \frac{17(4) - 13(5)}{60} = \frac{3}{60} \gt 0 \;\;\to\;\; \frac{17}{15} \gt \frac{13}{12}$$

which confirms that $$\sum_{i=1}^{30}\frac{1}{10 + i} \gt \frac{13}{12}$$.