how to find bounds of the series $\sum_{i=1}^n \frac1{4i-1}$

$$\sum_{i=1}^n \frac1{4i-1}$$

I know I have to integrate the function but from what to find lower and upper bound.

It sounds like you may be asking for a representation of $$\sum_{i=1}^n \frac{1}{4i-1}$$ as a Riemann sum estimate of two different integrals, one of which underestimates the true value of the integral (giving you an upper bound on your sum), and the other of which overestimates the true value of the integral (giving you a lower bound on your sum).

The right-hand Riemann sum approximation of $$\int_0^n \frac{1}{4x-1} dx$$ with $\Delta x = 1$ is $\sum_{i=1}^n \frac{1}{4i-1}$ and will underestimate the integral. However, the integrand is undefined at $x = 1/4$, and so you're better off looking at $$\frac{1}{3} + \sum_{i=2}^n \frac{1}{4i-1}$$ as an underestimate of $$\frac{1}{3} + \int_1^n \frac{1}{4x-1} dx.$$

Similarly, the left-hand Riemann sum approximation of $$\int_0^n \frac{1}{4(x+1)-1} dx$$ with $\Delta x = 1$ is $\sum_{i=1}^n \frac{1}{4i-1}$ and will overestimate the integral.

Drawing the Riemann sum approximations and the graphs of the functions will help to see this. Thus after putting it all together, we have

$$\int_0^n \frac{1}{4x+3} dx < \sum_{i=1}^n \frac{1}{4i-1} \leq \frac{1}{3} + \int_1^n \frac{1}{4x-1} dx,$$ with equality holding in the second case only when $n=1$.

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

Now apply the usual technique for obtaining a lower bound on the sum on the right.

• Is this supposed to be a joke? – TonyK Nov 22 '10 at 18:16
• Not at all. The only reason the standard underestimate couldn't be applied to the original problem was a singularity at 1/4. Pulling off one term removes the problem. – John D. Cook Nov 23 '10 at 0:00

You can write $$\frac{1}{4i-1} = \frac{1}{4i} + \frac{1}{4i(4i-1)}.$$ Therefore $$\sum_{i=1}^n \frac{1}{4i-1} = \frac{1}{4} \sum_{i=1}^n \frac{1}{i} + \sum_{i=1}^n \frac{1}{4i(4i-1)}.$$ The second term is bounded (by comparison with the convergent infinite series $\sum_{i=1}^\infty 1/i^2$), and the first term is familiar.