Key Concepts and Formulas

• Limit of a Sum as a Definite Integral (Riemann Sums): The limit of a sum can be converted into a definite integral using the formula: $\lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{n} f\left(\frac{r}{n}\right) = \int_0^1 f(x) dx$ This involves identifying the function $f(x)$ and the limits of integration by transforming the general term of the sum.
• Partial Fraction Decomposition: This technique is used to break down a complex rational function into a sum of simpler rational functions, making integration easier. For a term of the form $\frac{P(x)}{Q(x)}$, where $Q(x)$ can be factored, we express it as a sum of fractions with numerators that are constants or polynomials of lower degree than the corresponding factors in the denominator.
• Standard Integrals: Knowledge of basic integration formulas is required, such as:
  • $\int \frac{1}{a+x} dx = \log_e|a+x| + C$
  • $\int \frac{x}{a^2+x^2} dx = \frac{1}{2} \log_e(a^2+x^2) + C$
  • $\int \frac{1}{a^2+x^2} dx = \frac{1}{a} \arctan\left(\frac{x}{a}\right) + C$

Step-by-Step Solution

Step 1: Express the given sum in summation notation.

The given limit is: $S = \mathop {\lim }\limits_{n \to \infty } \left( {{{{n^2}} \over {({n^2} + 1^2)(n + 1)}} + {{{n^2}} \over {({n^2} + 2^2)(n + 2)}} + {{{n^2}} \over {({n^2} + 3^2)(n + 3)}} + \,\,....\,\, + \,\,{{{n^2}} \over {({n^2} + {n^2})(n + n)}}} \right)$ We can observe a pattern in the terms of the sum. The numerator is always $n^2$. The denominator consists of two factors. The first factor is of the form $(n^2 + r^2)$ and the second factor is of the form $(n+r)$, where $r$ takes integer values from $1$ to $n$. Therefore, the general term of the sum can be written as $\frac{n^2}{(n^2 + r^2)(n + r)}$. The sum can be expressed using summation notation as: $S = \mathop {\lim }\limits_{n \to \infty } \sum_{r=1}^{n} \frac{n^2}{(n^2 + r^2)(n + r)}$ Why this step? Converting the series into a summation form is the first crucial step to identify the general term and prepare it for transformation into a Riemann sum.

Step 2: Transform the general term to fit the Riemann Sum format.

The Riemann sum format requires the general term to be in the form $\frac{1}{n} f\left(\frac{r}{n}\right)$. Let's manipulate the general term $\frac{n^2}{(n^2 + r^2)(n + r)}$: Divide the numerator and the denominator by $n^3$ (since the highest power of $n$ in the denominator is $n^2 \cdot n = n^3$). $\frac{n^2}{(n^2 + r^2)(n + r)} = \frac{\frac{n^2}{n^3}}{\frac{(n^2 + r^2)(n + r)}{n^3}} = \frac{\frac{1}{n}}{\frac{n^2(1 + \frac{r^2}{n^2}) \cdot n(1 + \frac{r}{n})}{n^3}}$ $= \frac{\frac{1}{n}}{\frac{n^3(1 + (\frac{r}{n})^2)(1 + \frac{r}{n})}{n^3}} = \frac{1}{n} \cdot \frac{1}{\left(1 + \left(\frac{r}{n}\right)^2\right)\left(1 + \frac{r}{n}\right)}$ So, the limit becomes: $S = \mathop {\lim }\limits_{n \to \infty } \sum_{r=1}^{n} \frac{1}{n} \cdot \frac{1}{\left(1 + \left(\frac{r}{n}\right)^2\right)\left(1 + \frac{r}{n}\right)}$ Why this step? This transformation isolates the $\frac{1}{n}$ factor and expresses the rest of the term as a function of $\frac{r}{n}$, which is essential for applying the Riemann sum to definite integral conversion.

Step 3: Convert the limit of the sum into a definite integral.

Using the Riemann sum formula $\lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{n} f\left(\frac{r}{n}\right) = \int_0^1 f(x) dx$, we identify:

• $f(x) = \frac{1}{(1 + x^2)(1 + x)}$
• The limits of integration are from $0$ to $1$, as $r$ goes from $1$ to $n$, so $\frac{r}{n}$ goes from $\frac{1}{n} \to 0$ to $\frac{n}{n} = 1$. Thus, the limit of the sum is equivalent to the definite integral: $S = \int_0^1 \frac{1}{(1 + x^2)(1 + x)} dx$ Why this step? This step simplifies the problem from evaluating a limit of an infinite series to evaluating a definite integral, which is a standard calculus procedure.

Step 4: Evaluate the definite integral using Partial Fraction Decomposition.

We need to integrate $\frac{1}{(1 + x^2)(1 + x)}$. We use partial fraction decomposition: $\frac{1}{(1 + x^2)(1 + x)} = \frac{A}{1 + x} + \frac{Bx + C}{1 + x^2}$ Multiplying both sides by $(1 + x^2)(1 + x)$: $1 = A(1 + x^2) + (Bx + C)(1 + x)$ To find the constants $A$, $B$, and $C$:

• Let $x = -1$: $1 = A(1 + (-1)^2) + 0 \implies 1 = 2A \implies A = \frac{1}{2}$.
• Expand the equation: $1 = A + Ax^2 + Bx + Bx^2 + C + Cx$.
• Group terms by powers of $x$: $1 = (A+B)x^2 + (B+C)x + (A+C)$.
• Equating coefficients:
  • Coefficient of $x^2$: $0 = A + B$. Since $A = \frac{1}{2}$, $B = -\frac{1}{2}$.
  • Coefficient of $x$: $0 = B + C$. Since $B = -\frac{1}{2}$, $C = \frac{1}{2}$.
  • Constant term: $1 = A + C$. This is consistent with $A = \frac{1}{2}$ and $C = \frac{1}{2}$.

So, the partial fraction decomposition is: $\frac{1}{(1 + x^2)(1 + x)} = \frac{1/2}{1 + x} + \frac{-1/2 x + 1/2}{1 + x^2} = \frac{1}{2(1 + x)} + \frac{1 - x}{2(1 + x^2)}$ Now, we integrate term by term: $S = \int_0^1 \left( \frac{1}{2(1 + x)} + \frac{1 - x}{2(1 + x^2)} \right) dx$ $S = \frac{1}{2} \int_0^1 \frac{1}{1 + x} dx + \frac{1}{2} \int_0^1 \frac{1}{1 + x^2} dx - \frac{1}{2} \int_0^1 \frac{x}{1 + x^2} dx$ Evaluate each integral:

• $\int_0^1 \frac{1}{1 + x} dx = [\log_e|1 + x|]_0^1 = \log_e(2) - \log_e(1) = \log_e(2)$.
• $\int_0^1 \frac{1}{1 + x^2} dx = [\arctan(x)]_0^1 = \arctan(1) - \arctan(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4}$.
• $\int_0^1 \frac{x}{1 + x^2} dx$: Let $u = 1 + x^2$, so $du = 2x dx$, which means $x dx = \frac{1}{2} du$. When $x=0$, $u=1$. When $x=1$, $u=2$. $\int_1^2 \frac{1}{u} \cdot \frac{1}{2} du = \frac{1}{2} [\log_e|u|]_1^2 = \frac{1}{2} (\log_e(2) - \log_e(1)) = \frac{1}{2} \log_e(2)$.

Substitute these values back into the expression for $S$: $S = \frac{1}{2} (\log_e 2) + \frac{1}{2} \left(\frac{\pi}{4}\right) - \frac{1}{2} \left(\frac{1}{2} \log_e 2\right)$ $S = \frac{1}{2} \log_e 2 + \frac{\pi}{8} - \frac{1}{4} \log_e 2$ Combine the $\log_e 2$ terms: $S = \frac{\pi}{8} + \left(\frac{1}{2} - \frac{1}{4}\right) \log_e 2$ $S = \frac{\pi}{8} + \frac{1}{4} \log_e 2$ Why this step? Partial fraction decomposition simplifies the integrand into forms that are directly integrable using standard integration formulas. The evaluation of definite integrals then provides the numerical value of the original limit.

Common Mistakes & Tips

• Incorrectly identifying the general term: Carefully examine the pattern of the numerator and denominator to write the correct general term in summation notation.
• Errors in algebraic manipulation for Riemann sums: Ensure that the term is correctly divided by powers of $n$ to isolate $\frac{1}{n}$ and have a function solely of $\frac{r}{n}$.
• Mistakes in partial fraction decomposition or integration: Double-check the calculations for constants in partial fractions and the application of integration formulas, especially signs and factors of 2.

Summary

The problem was solved by first converting the given limit of a sum into a definite integral using the Riemann sum definition. This involved identifying the general term of the series, transforming it into the form $\frac{1}{n} f\left(\frac{r}{n}\right)$, and then applying the conversion formula. The resulting definite integral was then evaluated using partial fraction decomposition and standard integration techniques. The final result obtained is $\frac{\pi}{8} + \frac{1}{4} \log_e 2$.

The final answer is \boxed{{\pi \over 8} + {1 \over 4}{\log _e}2}.