What is $(3 n+1)+(3 n+2)+\ldots+(5 n)+(5 n+1)=$ equal to?

Question

I'm trying to prove this inequality$\frac{1}{2}<\frac{1}{3 n+1}+\frac{1}{3 n+2}+\ldots+\frac{1}{5 n}+\frac{1}{5 n+1}<\frac{2}{3}$, and i came across some difficulties with arithmetic progression $$3n+1+3n+2+\ldots +5n+5n+1=$$ In other cases, arithmetic progression formulas work just fine: $$ n+(n+1)+(n+2)+\ldots+2 n=(n+1)\left(\frac{n+2 n}{2}\right) $$ $2 n=n+(y-1)1 \rightarrow \Rightarrow \quad y=n+1 \\$ $$(n+1)+(n+2)+(n+2)+\ldots+(3 n+1)=(2 n+1)\left(\frac{4 n+2}{2}\right)$$ $3 n+1=n+1+(y-1)1\Rightarrow \quad-y=n-3 n-1 \Rightarrow y=2 n+1$

But in this case, it is not clear how I can determine what is the first member of this progression. Intuitively this may be understandable, but I would like to have a more strict definition. In other words, why $a_1 = 3n+1+3n+2$? $$(3 n+1)+(3 n+2)+\ldots+(5 n)+(5 n+1)=(4 n-1)\left(\frac{3 n+1+3 n+2+5 n+5 n+1}{2}\right)$$

$5 n+5 n+1=3 n+1+3 n+2+(y-1) 1 \Rightarrow 10 n+1=6 n+3+y-1 \\ -y=-10 n-1+6 n+3-1 \quad-y=-4 n+2-1 \quad y=4 n-1$

But $4n -1$ is wrong. As far as I know the correct answer is: $(4 n+1)\left(\frac{3 n+1+3 n+2+5 n+5 n+1}{2}\right)$. In this case, I got the correct answer and I can justify (left side) the inequality above using the arithmetic-harmonic mean inequality.

Answer 1

$\sum_{k=a}^bk=\frac{a+b}2(b-a+1).$ Applying this to $a=3n+1$ and $b=5n+1,$ you get $$(3n+1)+\dots+(5n+1)=(4n+1)(2n+1).$$ From this we deduce:

• your "As far as I know the correct answer is: $(4 n+1)\left(\frac{3 n+1+3 n+2+5 n+5 n+1}{2}\right)$" was wrong.
• your proof in comment for the lower bound $\frac12$ using the arithmetic-harmonic mean inequality is correct: $$\begin{align}\frac{1}{3 n+1}+\dots+\frac{1}{5 n+1}&\ge\frac{(2n+1)^2}{(3n+1)+\dots+(5n+1)}\\&=\frac{2n+1}{4n+1}\\&>\frac12.\end{align}$$

Answer 2

$$(3n+1)+...+(5n+1)=(3n+1)+...+(3n+n) + (4n+1)+...+(4n+n)+(5n+1)=(5n+1)+\sum^n_{k=1}(3n+k)+\sum^n_{k=1}(4n+k)=(5n+1)+\sum^n_{k=1}(7n+2k) = (5n+1)+\frac{n(16n+2)}{2}=8n^2+6n+1$$