# Induction with sum on left and right side.

• Tremoi
In summary, the given expression \sum_{k=1}^{n}(\frac{1}{2k-1} - \frac{1}{2k}) = \sum_{k=1}^{n}\frac{1}{k + n} holds for 1 \leq n and can be proven using induction.
Tremoi

## Homework Statement

Does
$\sum_{k=1}^{n}(\frac{1}{2k-1} - \frac{1}{2k}) = \sum_{k=1}^{n}\frac{1}{k + n}$
hold for $1 \leq n$

## The Attempt at a Solution

It holds for n = 1. I assume that it should be done with induction but I can't find a way actually compare the two sums to each other. I then had an idea about maybe putting each sum on a common denominator and prove that the both denominators and the both numerators are equal but that's not true so I don't really know where to start.
The left terms can be rewritten as $\frac{1}{2k(2k-1)}$ as well but that haven't really helped me either.

You can prove this using induction. Using the usual induction techniques, define $f(n) = \sum\limits_{k=1}^{k=n}\left(\frac{1}{2k-1} - \frac{1}{2k}\right)$. You want to show $f(n) = \sum\limits_{k=1}^{k=n} \frac{1}{k+n}$. You've already shown that this is true for $n=1$. Assuming the statement is true for $n$
$$f(n+1)\\ = \sum\limits_{k=1}^{k=n+1}\left(\frac{1}{2k-1} - \frac{1}{2k}\right) = f(n) + \frac{1}{2n+1} - \frac{1}{2n+2} \\ = \sum\limits_{k=1}^{k=n} \frac{1}{k+n}+ \frac{1}{2n+1} - \frac{1}{2n+2}$$
Now changing the variable in the sum $k \to k' = k-1$. The sum becomes
$$f(n+1) \\ = \sum\limits_{k'=0}^{k'=n-1} \frac{1}{k'+n+1}+ \frac{1}{2n+1} - \frac{1}{2n+2} \\ = \sum\limits_{k'=1}^{k'=n-1} \frac{1}{k'+n+1} + \frac{1}{n+1}+ \frac{1}{2n+1} - \frac{1}{2n+2}\\ = \sum\limits_{k'=1}^{k'=n-1} \frac{1}{k'+n+1}+ \frac{1}{2n+1} + \frac{1}{2n+2}\\ =\sum\limits_{k'=1}^{k'=n+1} \frac{1}{k'+n+1}$$

Thus, the statement is true!

## 1. What is induction with sum on left and right side?

Induction with sum on left and right side is a mathematical method used to prove that a statement or formula holds true for all natural numbers. It involves breaking down the formula into smaller parts and proving its validity for each part until it is proven for the entire range of natural numbers.

## 2. How is induction with sum on left and right side used?

Induction with sum on left and right side is typically used in proofs involving summation or series. It allows for a step-by-step approach to proving a formula or statement for all natural numbers, making it a powerful tool in mathematical reasoning and problem-solving.

## 3. What is the difference between induction with sum on left and right side and regular mathematical induction?

The main difference between induction with sum on left and right side and regular mathematical induction is that the former involves proving a formula or statement for all natural numbers by breaking it down into smaller parts and proving it for each part, whereas the latter involves proving a statement by assuming it is true for a specific number and then proving it for the next number.

## 4. What are the benefits of using induction with sum on left and right side?

Using induction with sum on left and right side allows for a systematic and organized approach to proving a formula or statement for all natural numbers. It also helps to break down complex problems into smaller, more manageable parts, making it easier to understand and solve.

## 5. Are there any limitations to using induction with sum on left and right side?

One limitation of using induction with sum on left and right side is that it can only be used to prove formulas or statements for natural numbers. It cannot be used for other types of numbers such as fractions or decimals. Additionally, some complex formulas may be difficult to break down and prove using this method.

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