Exercises and Solutions

1. Evaluate the limit:
   $$\underset{n\to \infty }{\lim }{\left(1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n}\right)}^{\frac{1}{n}}$$
   Solution:
   $$1\le \underset{n\to \infty }{\lim }{\left(1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n}\right)}^{\frac{1}{n}}\le \underset{n\to \infty }{\lim }{n}^{\frac{1}{n}}=1$$
   By the Squeeze Theorem:
   $$\underset{n\to \infty }{\lim }{\left(1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n}\right)}^{\frac{1}{n}}=1$$

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2. Evaluate the limit:
   $$\underset{n\to \infty }{\lim }\left(\frac{1}{n+\sqrt{1}}+\frac{1}{n+\sqrt{2}}+\cdots +\frac{1}{n+\sqrt{n}}\right)$$
   Solution:
   $$1=\underset{n\to \infty }{\lim }\sum _{i=1}^n\frac{1}{n+\sqrt{n}}\le \underset{n\to \infty }{\lim }\sum _{i=1}^n\frac{1}{n+\sqrt{i}}\le \underset{n\to \infty }{\lim }\sum _{i=1}^n\frac{1}{n}=1$$
   By the Squeeze Theorem:
   $$\underset{n\to \infty }{\lim }\sum _{i=1}^n\frac{1}{n+\sqrt{i}}=1$$

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3. Evaluate the limit:
   $$\underset{n\to \infty }{\lim }\sum _{k=n^2}^{(n+1{)}^2}\frac{1}{\sqrt{k}}$$
   Solution:
   $$\sum _{k=n^2}^{(n+1{)}^2}\frac{1}{\sqrt{(n+1{)}^2}}\le \sum _{k=n^2}^{(n+1{)}^2}\frac{1}{\sqrt{k}}\le \sum _{k=n^2}^{(n+1{)}^2}\frac{1}{\sqrt{n^2}}$$
   Taking the limit:
   $$\underset{n\to \infty }{\lim }2\le \underset{n\to \infty }{\lim }\sum _{k=n^2}^{(n+1{)}^2}\frac{1}{\sqrt{k}}\le \underset{n\to \infty }{\lim }\left(2+\frac{2}{n}\right)$$
   By the Squeeze Theorem:
   $$\underset{n\to \infty }{\lim }\sum _{k=n^2}^{(n+1{)}^2}\frac{1}{\sqrt{k}}=2$$

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4. Evaluate the limit:
   $$\underset{n\to \infty }{\lim }\frac{1\cdot 3\cdot 5\cdot \cdots \cdot (2n-1)}{2\cdot 4\cdot 6\cdot \cdots \cdot (2n)}$$
   Solution:
   Since:
   $$\frac{1}{2}\frac{(2n-2)!!}{(2n-1)!!}\le \frac{(2n-1)!!}{(2n)!!}\le \frac{(2n-2)!!}{(2n-1)!!}$$
   Then:
   $$\frac{1}{4n}\le {\left(\frac{(2n-1)!!}{(2n)!!}\right)}^2\le \frac{1}{2n}$$
   By the Squeeze Theorem, the limit exists:
   $$\underset{n\to \infty }{\lim }\frac{1\cdot 3\cdot 5\cdot \cdots \cdot (2n-1)}{2\cdot 4\cdot 6\cdot \cdots \cdot (2n)}=0$$

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5. Evaluate the limit:
   $$\underset{n\to \infty }{\lim }\frac{3n^2+4n-1}{n^2+1}$$
   Solution:
   $$\underset{n\to \infty }{\lim }\frac{3n^2+4n-1}{n^2+1}=3$$

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6. Evaluate the limit:
   $$\underset{n\to \infty }{\lim }\frac{n^3+2n^2-3n+1}{2n^3-n+3}$$
   Solution:
   $$\underset{n\to \infty }{\lim }\frac{n^3+2n^2-3n+1}{2n^3-n+3}=\frac{1}{2}$$

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7. Evaluate the limit:
   $$\underset{n\to \infty }{\lim }\frac{3^n+n^3}{3^{n+1}+(n+1{)}^3}$$
   Solution:
   $$\underset{n\to \infty }{\lim }\frac{3^n+n^3}{3^{n+1}+(n+1{)}^3}=\frac{1}{3}$$

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8. Evaluate the limit:
   $$\underset{n\to \infty }{\lim }\left(\sqrt[n]{n^2+1}-1\right)\sin \frac{n\pi }{2}$$
   Solution:
   $$\underset{n\to \infty }{\lim }\left(\sqrt[n]{n^2+1}-1\right)\sin \frac{n\pi }{2}=0$$

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9. Evaluate the limit:
   $$\underset{n\to \infty }{\lim }\sqrt{n}\left(\sqrt{n+1}-\sqrt{n}\right)$$
   Solution:
   $$\underset{n\to \infty }{\lim }\sqrt{n}\left(\sqrt{n+1}-\sqrt{n}\right)=\frac{1}{2}$$

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10. Evaluate the limit:
    $$\underset{n\to \infty }{\lim }\sqrt{n}\left(\sqrt[4]{n^2+1}-\sqrt{n+1}\right)$$
    Solution:
    $$\underset{n\to \infty }{\lim }\sqrt{n}\left(\sqrt[4]{n^2+1}-\sqrt{n+1}\right)=-\frac{1}{2}$$

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11. Evaluate the limit:
    $$\underset{n\to \infty }{\lim }\sqrt[n]{\frac{1}{n!}}$$
    Solution:
    $$\underset{n\to \infty }{\lim }\sqrt[n]{\frac{1}{n!}}=0$$

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12. Evaluate the limit:
    $$\underset{n\to \infty }{\lim }\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\cdots \left(1-\frac{1}{n^2}\right)$$
    Solution:
    $$\underset{n\to \infty }{\lim }\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\cdots \left(1-\frac{1}{n^2}\right)=\frac{1}{2}$$

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13. Evaluate the limit:
    $$\underset{n\to \infty }{\lim }\sqrt[n]{n\ln n}$$
    Solution:
    $$\underset{n\to \infty }{\lim }\sqrt[n]{n\ln n}=0$$

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14. Evaluate the limit:
    $$\underset{n\to \infty }{\lim }\left(\frac{1}{2}+\frac{3}{2^2}+\cdots +\frac{2n-1}{2^n}\right)$$
    Solution:
    $$\underset{n\to \infty }{\lim }\left(\frac{1}{2}+\frac{3}{2^2}+\cdots +\frac{2n-1}{2^n}\right)=3$$

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15. Given ${\lim }_{n\to \infty }{a}_n=a$ and ${\lim }_{n\to \infty }{b}_n=b$, prove:
    $$\underset{n\to \infty }{\lim }\frac{a_1{b}_n+a_2{b}_{n-1}+\cdots +a_n{b}_1}{n}=ab$$
    Proof:
    Let $a_n=a+{\alpha }_n$ and $b_n=b+{\beta }_n$. Then:
    $$\frac{{\sum }_{i=1}^n{a}_i{b}_{n+1-i}}{n}=ab+\frac{b}{n}\sum _{i=1}^n{\alpha }_i+\frac{a}{n}\sum _{i=1}^n{\beta }_i+\frac{1}{n}\sum _{i=1}^n{\alpha }_i{\beta }_{n-i+1}$$
    Let $|{\alpha }_n|,|{\beta }_n|\le M$. For $n>N_1$ and $n>N_2$:
    $$\left|\frac{1}{n}\sum _{i=1}^n{\alpha }_i{\beta }_{n-i+1}\right|<\left|\frac{1}{n}\sum _{i=1}^N{\alpha }_i{\beta }_{n-i+1}\right|+\left|\frac{1}{n}\sum _{i=N+1}^n{\beta }_{n-i+1}\right|\epsilon <\frac{N{M}^2}{n}+M\epsilon$$
    Therefore:
    $$\underset{n\to \infty }{\lim }\frac{{\sum }_{i=1}^n{a}_i{b}_{n+1-i}}{n}=ab$$