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# Determine the convergence or divergence of f(n) = (n+1)1/2 – n1/2

Consider the function defined by

Determine whether the sequence converges or diverges, and if it converges find the limit.

First, we use multiply the numerator and denominator by ,

Then, since we have

Since we know from property (10.9) on page 380 of Apostol that

we then have

Hence, by the squeeze theorem we have

Therefore the sequence converges with limit 0.

# Determine the convergence or divergence of f(n) = (1 + (-1)n) / n

Consider the function defined by

Determine whether the sequence converges or diverges, and if it converges find the limit.

To show is convergent we will use the squeeze theorem to determine its limit. First we note

for all (since if is odd and if is even). Since

by the squeeze theorem we have

# In two different ways, prove the limit as x tends to 0 of (log (1+x))/x is 1

The following limit equation is valid:

Prove this in the following two ways:

1. Using the definition of the derivative ;
2. Using the previous exercise (Section 6.9, Exercise #28).

1. Proof. First, we know from Theorem 6.1 part (b) (page 229 of Apostol) that

Therefore, . Then, we recall the definition of the derivative of a function

So, using the definition of the derivative and the fact that for we have ,

Evaluating at we have

since the variable name is unimportant. Thus, we have established the requested formula,

2. Proof. From the previous exercise (Section 6.9, Exercise #28) we know

If then as well, so the inequalities still hold with ,

Multiplying all of the terms by (since we may do this without reversing the inequalities),

Since we then have,

Finally, since

we apply the squeeze theorem (Theorem 3.3, page 133 of Apostol) to conclude,

# Prove a squeeze-like property for reals.

If and for all , then .

Proof. Suppose otherwise, that for all , but . Then, we have and implies , so is a positive real number. Then let (since this is a valid choice of ). By assumption we then have (since for all ), but this is a contradiction since (equality is symmetric) implies . Therefore, we must have