Series de números reales

Series alternadas

Series de términos no-negativos

Convergencia absoluta y condicionaol

Criterios de Dirichlet y de Abel

Reordenación de series

Producto de Cauchy de dos series

---

To prove the formula for the sum of a finite geometric progression, we want to show that:

a + ar + ar^2 + \dots + ar^{n-1} = \frac{a(1 - r^n)}{1 - r},

where $( a )$ and $( r )$ are real numbers, and $( r \neq 1 )$.

Step-by-Step Proof

1. Define the Sum of the Series:

   Let $( S )$ be the sum of the first $( n )$ terms of the geometric series:

   S = a + ar + ar^2 + \dots + ar^{n-1}.

2. Express $( S )$ as a Product:

   Notice that each term in $( S )$ is multiplied by a successive power of $( r )$. We can factor out $( a )$ to rewrite $( S )$ as:

   S = a(1 + r + r^2 + \dots + r^{n-1}).

3. Multiply $( S )$ by $( r )$:

   Consider $( rS )$, which is obtained by multiplying each term of $( S )$ by $( r )$:

   rS = ar + ar^2 + ar^3 + \dots + ar^n.

Notice that this new series $( rS )$ is similar to $( S )$ but shifted by one term. We can write this as:

   rS = ar + ar^2 + ar^3 + \dots + ar^{n-1} + ar^n.

4. Subtract $( rS )$ from $( S )$:

   Now, subtract $( rS )$ from $( S )$ to eliminate all terms except the first term $( a )$ and the last term $( -ar^n )$:

   S - rS = (a + ar + ar^2 + \dots + ar^{n-1}) - (ar + ar^2 + \dots + ar^{n-1} + ar^n).

This simplifies to:

   S - rS = a - ar^n.

5. Factor Out $( S )$:

   Now factor $( S )$ out on the left side:

   S(1 - r) = a - ar^n.

6. Solve for $( S )$:

   Divide both sides by $( 1 - r )$ (noting that $( r \neq 1 )$ to avoid division by zero):

   S = \frac{a(1 - r^n)}{1 - r}.

Conclusion

We have shown that the sum of the first $( n )$ terms of a geometric series is:

\boxed{a + ar + ar^2 + \dots + ar^{n-1} = \frac{a(1 - r^n)}{1 - r}},

as required.

---

To prove that for all positive integers $( n )$,

\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n(n + 1)} = \frac{n}{n + 1},

we will proceed by mathematical induction.

Step 1: Base Case

First, we verify the formula for $( n = 1 )$.

For $( n = 1 )$:

• The left side of the equation is simply $( \frac{1}{1 \cdot 2} = \frac{1}{2} )$.
• The right side of the equation, with $( n = 1 )$, becomes:

  \frac{1}{1 + 1} = \frac{1}{2}.

Since both sides are equal, the formula holds for $( n = 1 )$. Thus, the base case is true.

Step 2: Inductive Hypothesis

Assume that the formula holds for some positive integer $( k )$; that is, assume:

\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{k(k + 1)} = \frac{k}{k + 1}.

This assumption is called the inductive hypothesis.

Step 3: Inductive Step

We need to prove that if the formula holds for $( n = k )$, then it also holds for $( n = k + 1 )$. In other words, we need to show:

\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{k(k + 1)} + \frac{1}{(k + 1)(k + 2)} = \frac{k + 1}{k + 2}.

Proof of the Inductive Step

Starting from the left side of the equation for $( n = k + 1 )$, we have:

\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{k(k + 1)} + \frac{1}{(k + 1)(k + 2)}.

By the inductive hypothesis, we know that:

\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{k(k + 1)} = \frac{k}{k + 1}.

Substituting this into our expression, we get:

\frac{k}{k + 1} + \frac{1}{(k + 1)(k + 2)}.

Now, let’s combine these two terms over a common denominator, $( (k + 1)(k + 2) )$:

\frac{k}{k + 1} + \frac{1}{(k + 1)(k + 2)} = \frac{k(k + 2) + 1}{(k + 1)(k + 2)}.

Expanding the numerator:

= \frac{k^2 + 2k + 1}{(k + 1)(k + 2)}.

Now, factor the numerator as a perfect square:

= \frac{(k + 1)^2}{(k + 1)(k + 2)}.

Cancel $( k + 1 )$ in the numerator and denominator:

= \frac{k + 1}{k + 2}.

This is exactly the right side of our desired equation for $( n = k + 1 )$:

\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{(k + 1)(k + 2)} = \frac{k + 1}{k + 2}.

Conclusion

Since we have shown that:

1. The formula holds for the base case $( n = 1 )$,
2. If the formula holds for $( n = k )$, then it also holds for $( n = k + 1 )$,

we conclude by the principle of mathematical induction that the formula is true for all positive integers $( n )$:

\boxed{\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n(n + 1)} = \frac{n}{n + 1}}.