Question

Use differentials to show that, for very large n , $\frac{\mathbf{1}}{{\left(n+1\right)}^{\mathbf{3}}}\mathbf{-}\frac{\mathbf{1}}{{\mathbf{n}}^{\mathbf{3}}}\mathbf{\equiv }\frac{\mathbf{3}}{{\mathbf{n}}^{\mathbf{4}}}$.

Short answer

The answer is:$\frac{1}{{(\mathrm{n}+1)}^3}-\frac{1}{{\mathrm{n}}^3}\equiv \frac{3}{{\mathrm{n}}^4}$

Step-by-step solution

Explanation of Solution

As the provided equation is$\frac{1}{{(n+1)}^3}-\frac{1}{n^3}\cong \frac{3}{n^4}$ .

Approximation by differentials

This method is based on the derivatives of functions whose values must be calculated at certain locations, as the name implies.

Consider a function $\mathbf{y}\mathbf{=}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$, find the value of a function$\mathbf{y}\mathbf{=}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$when$\mathit{x}\mathbf{=}\mathit{x}\mathbf{\text{'}}$.

As an example, the derivative of a function $\mathit{y}\mathbf{=}\mathit{f}\left(x\right)$with respect to xwill be employed.

$\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{=}\left(f\left(x\right)\right)$is Change in ywith respect to change in xas $\mathit{d}\mathit{x}\mathbf{\Rightarrow }\mathbf{0}$.

If the value of $x=x\text{'}$from a value of x near it, such that the difference in the two values, dx , is vanishingly small, one can derive the change in the value of the function $y=f\left(x\right)$corresponding to the change dx in x . In practice, however, the concept of vanishingly small is not possible.

Calculation

The stated equation is $\frac{1}{{(n+1)}^3}-\frac{1}{n^3}\cong \frac{3}{n^4}$.

If $f\left(x\right)=\frac{1}{x^3}$is defined, the necessary differences are $∆f=f\left(n+1\right)-f\left(n\right)$. However, because $∆f$is close to df (for extremely large n ), the above differences can be rewritten using differentials:

$$\begin{gathered} \mathrm{df}=\frac{\mathrm{df}}{\mathrm{dx}}\mathrm{dx} \\ =\mathrm{d}\left(\frac{1}{{\mathrm{x}}^3}\right) \\ =\frac{-3}{{\mathrm{x}}^4}\mathrm{dx} \end{gathered}$$

With$x=n$and$dx=1$

$\begin{array}{l}\mathrm{df}=\frac{-3}{{\mathrm{n}}^4}\times 1\\ =\frac{-3}{{\mathrm{n}}^4}\end{array}$

Hence,$\frac{1}{{(n+1)}^3}-\frac{1}{n^3}\equiv \frac{3}{n^4}$