1. Condesing Summations

Okay, I have the following:

$
\sum_{n=1}^{4}\frac{1}{n} + 1 + \sum_{n=1}^{3}\frac{1}{n} + 1 + \sum_{n=1}^{2}\frac{1}{n} + 1 + \sum_{n=1}^{1}\frac{1}{n}+1
$

I want to repeat this N times, so at any N I can find the total, so I suppose:

$
\sum_{n=1}^{N}\frac{1}{n}+\sum_{n=1}^{N-1}\frac{1}{n}+....+\sum_{n=1}^{2}\frac{1}{n}+\sum_ {n=1}^{1}\frac{1}{n}+N
$

I want to write this in a single condensed sum (well N could be added on the end) or something that is at least simple, could someone help me? thanks

2. This is an interesting problem. From my thinking on it, I don't think it's possible to write this only using one sigma. The reason I think so is that you have too many things to keep track of and not enough variables in your current explanation. The total sum you seek is defined by other sums, so you must either define the smaller sums with another sigma or find a nice pattern in the sums that allows you to express the current total in terms of N.

I thought of it like this:

$S = \left([N*\frac{1}{n}]+[(N-1)*\frac{1}{n+1}]...[(N-(N-1))*\frac{1}{N}] \right)+N$ and 1 < n < N.

We must find a way to relate the increasing value of "n" and the decreasing value of "N" in terms of one variable. I can't think of a way to do it. Maybe some of these thoughts will be useful or someone else can solve it.

3. Originally Posted by jarny
Okay, I have the following:

$
\sum_{n=1}^{4}\frac{1}{n} + 1 + \sum_{n=1}^{3}\frac{1}{n} + 1 + \sum_{n=1}^{2}\frac{1}{n} + 1 + \sum_{n=1}^{1}\frac{1}{n}+1
$

I want to repeat this N times, so at any N I can find the total, so I suppose:

$
\sum_{n=1}^{N}\frac{1}{n}+\sum_{n=1}^{N-1}\frac{1}{n}+....+\sum_{n=1}^{2}\frac{1}{n}+\sum_ {n=1}^{1}\frac{1}{n}+N
$

I want to write this in a single condensed sum (well N could be added on the end) or something that is at least simple, could someone help me? thanks
I think the sum that you seek is $(N+1)\sum_{n=1}^N \frac{1}{n}$

4. Originally Posted by Jameson
This is an interesting problem. From my thinking on it, I don't think it's possible to write this only using one sigma. The reason I think so is that you have too many things to keep track of and not enough variables in your current explanation. The total sum you seek is defined by other sums, so you must either define the smaller sums with another sigma or find a nice pattern in the sums that allows you to express the current total in terms of N.

I thought of it like this:

$S = \left([N*\frac{1}{n}]+[(N-1)*\frac{1}{n+1}]...[(N-(N-1))*\frac{1}{N}] \right)+N$ and 1 < n < N.

We must find a way to relate the increasing value of "n" and the decreasing value of "N" in terms of one variable. I can't think of a way to do it. Maybe some of these thoughts will be useful or someone else can solve it.

$\sum_{n=1}^{N}\frac{N-n+1}{n}$

This is similar to Danny's, except that mine is $\left((N+1)\sum_{n=1}^N\frac{1}{n}\right)-N$

Only one can be right; which is it?

5. Actually I was just pming Jameson, I got the same thing danny got actually hahaha I'm having trouble seeing why that is though. I was doing guess and check on my calculator and ran into the answer. I want to know why that is though :/

6. Hello, jarny!

Okay, I have the following:

$\sum_{n=1}^{4}\frac{1}{n} + 1 + \sum_{n=1}^{3}\frac{1}{n} + 1 + \sum_{n=1}^{2}\frac{1}{n} + 1 + \sum_{n=1}^{1}\frac{1}{n}+1
$

I want to repeat this $N$ times, so at any $N$, I can find the total, so I suppose:

$\sum_{n=1}^{N}\frac{1}{n}+\sum_{n=1}^{N-1}\frac{1}{n}+ \hdots +\sum_{n=1}^{2}\frac{1}{n}+\sum_{n=1}^{1}\frac{1}{ n}+N
$

I want to write this in a single condensed sum (N could be added on the end) .
I wrote out the general list . . . and added.

$\sum^1_{k=1}\frac{1}{k} \;=\;\frac{1}{1}$

$\sum^2_{k=1}\frac{1}{k} \;=\;\frac{1}{1} + \frac{1}{2}$

$\sum^3_{k=1}\frac{1}{k} \;=\;\frac{1}{1} + \frac{1}{2} + \frac{1}{3}$

$\sum^4_{k=1}\frac{1}{k} \;=\;\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}$

. . $\vdots$. . . . . . . . $\vdots$

$\sum^{n-1}_{k=1}\frac{1}{k} \;=\;\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \hdots + \frac{1}{n-1}$

$\sum^n_{k=1}\frac{1}{k} \;=\;\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \hdots + \frac{1}{n-1} + \frac{1}{n}$

$\text{Sum} \;=\;n\left(\frac{1}{1}\right) + (n-1)\left(\frac{1}{2}\right) + (n-2)\left(\frac{1}{3}\right) + \hdots + 2\left(\frac{1}{n-1}\right) + 1\left(\frac{1}{n}\right)$

. . . $= \;\frac{n}{1} + \frac{n-1}{2} + \frac{n-2}{3} + \frac{n-3}{4} + \hdots + \frac{2}{n-1} + \frac{1}{n}$

. . . $= \;\sum^n_{k=1}\frac{n-k+1}{k}$

7. Soroban, Your sum comes up with a different result...I'm trying to figure out what is wrong though :/

8. Okay, Your way is much much easier, I think you just forgot to add n at the end. Thank you very much!!!