# Proving 1 + 1/√2 +...+ 1/√n < 2√n Using Mathematical Induction

• trap101
In summary: You get2 \sqrt{k+1} - 2 \sqrt{k} = 2 \sqrt{k+1} - 2 \sqrt{k+1} + 2 \sqrt{k}= 2 \sqrt{k} = 2 \sqrt{k} \cdot 1. Now note that1 = \frac{k+1}{k+1} = \frac{k+1}{k+1} \cdot 1 = \frac{k+1}{k+1} \cdot \frac{\sqrt{k}}{\sqrt{k}} = \frac{\sqrt{k^2}}{\sqrt{k(k+1)}} = \frac{\sqrt{k^2 + k}}{\sqrt{k(k+1)}}
trap101
Prove that:

1 + 1/√2 +...+ 1/√n < 2√n

Work:

So I've done the base case of n = 1, and I've set up the Indutive hypothesis as assuming n=k as 1 + 1/√2 +...+ 1/√k < 2√k

and for the inductive step:

1 + 1/√2 +...+ 1/√k + 1/√(k+1) < 2√k + 1/√(k+1)

now I'm having trouble trying to manipulate 2√k + 1/√(k+1) to get 2√(k+1) alone. I tried a conjugate of 1/[2 √(k+1) + 2√k] [2 √(k+1) - 2√k / [2 √(k+1) - 2√k]...the closest thing I got was 2√(k+1) / 4.

ANy suggestions?

Here is a hint which should be useful:
$$4(k^2 + k) < 4k^2 + 4k + 1 = (2k + 1)^2$$
Taking the square root of both sides, which is OK because both sides are positive, we get
$$2\sqrt{k^2 + k} < 2k + 1$$
See if you can manipulate the right hand side of your inequality into a form where you can apply this.

Last edited:
trap101 said:
Prove that:

1 + 1/√2 +...+ 1/√n < 2√n

Work:

So I've done the base case of n = 1, and I've set up the Indutive hypothesis as assuming n=k as 1 + 1/√2 +...+ 1/√k < 2√k

and for the inductive step:

1 + 1/√2 +...+ 1/√k + 1/√(k+1) < 2√k + 1/√(k+1)

now I'm having trouble trying to manipulate 2√k + 1/√(k+1) to get 2√(k+1) alone. I tried a conjugate of 1/[2 √(k+1) + 2√k] [2 √(k+1) - 2√k / [2 √(k+1) - 2√k]...the closest thing I got was 2√(k+1) / 4.

ANy suggestions?

So, you want to show $$2 \sqrt{k+1} - 2 \sqrt{k} \geq \frac{1}{\sqrt{k+1}}.$$ Re-write the LHJS using the type of trick you applied above.

## 1. What is simple mathematical induction?

Simple mathematical induction is a method used to prove that a statement is true for all natural numbers. It involves two steps: the base case, where the statement is proven to be true for the first natural number, and the inductive step, where it is shown that if the statement is true for one natural number, it is also true for the next natural number.

## 2. How is simple mathematical induction different from strong mathematical induction?

Simple mathematical induction only requires the base case and the inductive step, while strong mathematical induction also allows for multiple base cases and requires a stronger inductive step.

## 3. What types of statements can be proven using simple mathematical induction?

Simple mathematical induction can be used to prove statements about natural numbers, such as properties of sequences or sums, divisibility, and inequalities.

## 4. Can simple mathematical induction be used to prove statements for infinite sets?

No, simple mathematical induction can only be used to prove statements for finite sets, specifically natural numbers. For infinite sets, other methods such as strong mathematical induction or transfinite induction may be used.

## 5. What are some common mistakes to avoid when using simple mathematical induction?

Some common mistakes to avoid when using simple mathematical induction include assuming the statement is true for all natural numbers without properly proving it, using incorrect or insufficient base cases, and using invalid or incomplete inductive steps.

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