Evaluate $\displaystyle \lim_{x \to \infty}\dfrac{1}{{{n}^{4}}}\left( 1\sum\limits_{k=1}^{n}{k}+2\sum\limits_{k=1}^{n-1}{k}+3\sum\limits_{k=1}^{n-2}{k...+n\cdot 1} \right)$ \[\]
Answer
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Hint: We can write the summation $1\sum\limits_{k=1}^{n}{k}+2\sum\limits_{k=1}^{n-1}{k}+3\sum\limits_{k=1}^{n-2}{k+...+n\cdot 1}$ as $\sum\limits_{k=1}^{n}{\left( k\sum\limits_{r=1}^{r=n-k+1}{r} \right)}$ where the counter $k$ runs for the multipliers outside $1,2,3...n$ and $r$ is the counter replacing $k$ which runs for the summation in each term . We expand $\sum\limits_{k=1}^{n}{\left( k\sum\limits_{r=1}^{r=n-k+1}{r} \right)}$ using the formula of sum of first $n$ terms $\left( \sum\limits_{k=1}^{n}{k=\dfrac{n\left( n+1 \right)}{2}} \right)$ , squared $n$ terms $\left( \sum\limits_{k=1}^{n}{{{k}^{2}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}} \right)$ and cubed $n$ terms $\left( \sum\limits_{k=1}^{n}{{{k}^{3}}=\dfrac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4}} \right)$ . We find an expression free of $k,r$ and put in the given function in $n$ to evaluate. \[\]
Complete step by step answer:
We know that limiting value for any real valued single variable function $f\left( x \right)$ when the variable $x$ approaches to real number $a$ in the domain $f\left( x \right)$ is denoted by
\[\displaystyle \lim_{x \to a}f\left( x \right)=L\]
Here $L$ is called the limit of the function. We know that $\displaystyle \lim_{x \to \infty}\dfrac{1}{x}=0$
We have the given function in $n$ to evaluate the limit is
\[\displaystyle \lim_{x \to \infty}\dfrac{1}{{{n}^{4}}}\left( 1\sum\limits_{k=1}^{n}{k}+2\sum\limits_{k=1}^{n-1}{k}+3\sum\limits_{k=1}^{n-2}{k...+n\cdot 1} \right)\]
Let us observe the terms inside the bracket $1\sum\limits_{k=1}^{n}{k}+2\sum\limits_{k=1}^{n-1}{k}+3\sum\limits_{k=1}^{n-2}{k+...+n\cdot 1}$ which is also a summation of $n$ number of terms with each term containing a summation. We take counter $k$ for the multipliers which runs from $k=1$ to $k=n$ for the whole sequence $1\sum\limits_{k=1}^{n}{k},2\sum\limits_{k=1}^{n-1}{k},3\sum\limits_{k=1}^{n-2}{k,...,n\cdot 1}$. \[\]
Let us find the replace the counter the summations $k$ with $r$ and find the counter ${{r}^{\text{th}}}$ term of the sequence $1\sum\limits_{k=1}^{n}{k},2\sum\limits_{k=1}^{n-1}{k},3\sum\limits_{k=1}^{n-2}{k,...,n\cdot 1}$. We see that the first term is added up from $k=1$ to $k=n-1+1$ , the second term is added from $k=1$ to $k=n-2+1$, the third term is added from $k=1$ to $k=n-3+1$ and so on. The ${{r}^{\text{th}}}$ will be added from $r=1$ to $r=n-k+1$. So we have the transformed sum as ,
\[1\sum\limits_{k=1}^{n}{k}+2\sum\limits_{k=1}^{n-1}{k}+3\sum\limits_{k=1}^{n-2}{k...+n\cdot 1}=\sum\limits_{k=1}^{n}{\left( k\sum\limits_{r=1}^{r=n-k+1}{r} \right)}=\sum\limits_{k=1}^{n}{k\left( \sum\limits_{r=1}^{r=n-k+1}{r} \right)}\]
We use the sum of first $n$ terms formula and get ,
\[\begin{align}
& \sum\limits_{k=1}^{n}{k\left( \sum\limits_{r=1}^{r=n-k+1}{r} \right)}=\sum\limits_{k=1}^{n}{k\left( 1+2+3...+n-k+1 \right)} \\
& \Rightarrow \sum\limits_{k=1}^{n}{k\left( \sum\limits_{r=1}^{r=n-k+1}{r} \right)}=\sum\limits_{k=1}^{n}{k\dfrac{\left( n-k+1 \right)\left( n-k+2 \right)}{2}} \\
& \Rightarrow \sum\limits_{k=1}^{n}{k\left( \sum\limits_{r=1}^{r=n-k+1}{r} \right)}=\sum\limits_{k=1}^{n}{k\dfrac{{{n}^{2}}+{{k}^{2}}-\left( 2n+3 \right)k+3n+2}{2}} \\
& \Rightarrow \sum\limits_{k=1}^{n}{k\left( \sum\limits_{r=1}^{r=n-k+1}{r} \right)}=\sum\limits_{k=1}^{n}{k\dfrac{{{n}^{2}}+3n+2}{2}}-\sum\limits_{k=1}^{n}{k\times k\dfrac{2n+3}{2}}+\dfrac{1}{2}\sum\limits_{k=1}^{n}{k\times {{k}^{2}}} \\
\end{align}\]
We see that the terms involving $n$ are constant and they can be taken outside of the summation. So we have
\[\Rightarrow \sum\limits_{k=1}^{n}{k\left( \sum\limits_{r=1}^{r=n-k+1}{r} \right)}=\dfrac{{{n}^{2}}+3n+2}{2}\sum\limits_{k=1}^{n}{k}-\dfrac{2n+3}{2}\sum\limits_{k=1}^{n}{{{k}^{2}}}+\dfrac{1}{2}\sum\limits_{k=1}^{n}{{{k}^{3}}}\]
We use formulas for the sum of first $n$ terms , the sum of first $n$ squared terms and the sum of first $n$ cubed terms and get,
\[\begin{align}
& =\dfrac{{{n}^{2}}+3n+2}{2}\left( \dfrac{n\left( n+1 \right)}{2} \right)-\dfrac{2n+3}{2}\left( \dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6} \right)+\dfrac{1}{2}{{\left( \dfrac{n\left( n+1 \right)}{2} \right)}^{2}} \\
& =\dfrac{n\left( n+1 \right)}{2}\left( \dfrac{{{n}^{2}}+3n+2}{2}-\dfrac{\left( 2n+3 \right)\left( 2n+1 \right)}{6}+\dfrac{n\left( n+1 \right)}{4} \right) \\
& =\dfrac{n\left( n+1 \right)}{2}\left( \dfrac{6{{n}^{2}}+18n+12}{12}-\dfrac{8{{n}^{2}}+16n+6}{12}+\dfrac{3{{n}^{2}}+3n}{12} \right) \\
& =\dfrac{n\left( n+1 \right)\left( {{n}^{2}}+5n+6 \right)}{24} \\
\end{align}\]
We replace ${{n}^{2}}+5n+6$ with its factorization $\left( n+2 \right)\left( n+3 \right)$ in the above result and get
\[\sum\limits_{k=1}^{n}{k\left( \sum\limits_{r=1}^{r=n-k+1}{r} \right)}=\dfrac{n\left( n+1 \right)\left( n+2 \right)\left( n+3 \right)}{24}\]
Now we can put the above result which is free of $k$ in the given limit and get,
\[\begin{align}
& \displaystyle \lim_{x \to \infty}\dfrac{1}{{{n}^{4}}}\left( 1\sum\limits_{k=1}^{n}{k}+2\sum\limits_{k=1}^{n-1}{k}+3\sum\limits_{k=1}^{n-2}{k...+n\cdot 1} \right)=\displaystyle \lim_{x \to \infty}\dfrac{1}{{{n}^{4}}}\sum\limits_{k=1}^{n}{k\left( \sum\limits_{r=1}^{r=n-k+1}{r} \right)} \\
& =\displaystyle \lim_{x \to \infty}\dfrac{1}{{{n}^{4}}}\left( \dfrac{n\left( n+1 \right)\left( n+2 \right)\left( n+3 \right)}{24} \right) \\
\end{align}\]
Let us take $n$ common from $\left( n+1 \right),\left( n+2 \right),\left( n+3 \right)$ and get
\[\begin{align}
& =\displaystyle \lim_{x \to \infty}\dfrac{1}{{{n}^{4}}}\left( \dfrac{{{n}^{4}}\left( 1+\dfrac{1}{n} \right)\left( 1+\dfrac{2}{n} \right)\left( n+\dfrac{3}{n} \right)}{24} \right) \\
& =\displaystyle \lim_{x \to \infty}\dfrac{\left( 1+\dfrac{1}{n} \right)\left( 1+\dfrac{2}{n} \right)\left( n+\dfrac{3}{n} \right)}{24} \\
\end{align}\]
Now e use the law of multiplication of limits and get
\[\begin{align}
& =\dfrac{\displaystyle \lim_{x \to \infty}\left( 1+\dfrac{1}{n} \right)\displaystyle \lim_{x \to \infty}\left( 1+\dfrac{2}{n} \right)\displaystyle \lim_{x \to \infty}\left( n+\dfrac{3}{n} \right)}{24} \\
& \Rightarrow \displaystyle \lim_{x \to \infty}\dfrac{1}{{{n}^{4}}}\left( 1\sum\limits_{k=1}^{n}{k}+2\sum\limits_{k=1}^{n-1}{k}+3\sum\limits_{k=1}^{n-2}{k...+n\cdot 1} \right)=\dfrac{1\times 1\times 1}{24}=\dfrac{1}{24} \\
\end{align}\]
Note: We know that the limit exists for real valued single variable function $f\left( x \right)$ at any point $x=a$ if and only if Left hand limit$\left( \displaystyle \lim_{x \to {{a}^{-}}}f\left( x \right) \right)=$right hand limit$\left( \displaystyle \lim_{x \to {{a}^{+}}}f\left( x \right) \right)$. We note that we have only evaluated only right hand limit because the function is continuous everywhere in $R.$
Complete step by step answer:
We know that limiting value for any real valued single variable function $f\left( x \right)$ when the variable $x$ approaches to real number $a$ in the domain $f\left( x \right)$ is denoted by
\[\displaystyle \lim_{x \to a}f\left( x \right)=L\]
Here $L$ is called the limit of the function. We know that $\displaystyle \lim_{x \to \infty}\dfrac{1}{x}=0$
We have the given function in $n$ to evaluate the limit is
\[\displaystyle \lim_{x \to \infty}\dfrac{1}{{{n}^{4}}}\left( 1\sum\limits_{k=1}^{n}{k}+2\sum\limits_{k=1}^{n-1}{k}+3\sum\limits_{k=1}^{n-2}{k...+n\cdot 1} \right)\]
Let us observe the terms inside the bracket $1\sum\limits_{k=1}^{n}{k}+2\sum\limits_{k=1}^{n-1}{k}+3\sum\limits_{k=1}^{n-2}{k+...+n\cdot 1}$ which is also a summation of $n$ number of terms with each term containing a summation. We take counter $k$ for the multipliers which runs from $k=1$ to $k=n$ for the whole sequence $1\sum\limits_{k=1}^{n}{k},2\sum\limits_{k=1}^{n-1}{k},3\sum\limits_{k=1}^{n-2}{k,...,n\cdot 1}$. \[\]
Let us find the replace the counter the summations $k$ with $r$ and find the counter ${{r}^{\text{th}}}$ term of the sequence $1\sum\limits_{k=1}^{n}{k},2\sum\limits_{k=1}^{n-1}{k},3\sum\limits_{k=1}^{n-2}{k,...,n\cdot 1}$. We see that the first term is added up from $k=1$ to $k=n-1+1$ , the second term is added from $k=1$ to $k=n-2+1$, the third term is added from $k=1$ to $k=n-3+1$ and so on. The ${{r}^{\text{th}}}$ will be added from $r=1$ to $r=n-k+1$. So we have the transformed sum as ,
\[1\sum\limits_{k=1}^{n}{k}+2\sum\limits_{k=1}^{n-1}{k}+3\sum\limits_{k=1}^{n-2}{k...+n\cdot 1}=\sum\limits_{k=1}^{n}{\left( k\sum\limits_{r=1}^{r=n-k+1}{r} \right)}=\sum\limits_{k=1}^{n}{k\left( \sum\limits_{r=1}^{r=n-k+1}{r} \right)}\]
We use the sum of first $n$ terms formula and get ,
\[\begin{align}
& \sum\limits_{k=1}^{n}{k\left( \sum\limits_{r=1}^{r=n-k+1}{r} \right)}=\sum\limits_{k=1}^{n}{k\left( 1+2+3...+n-k+1 \right)} \\
& \Rightarrow \sum\limits_{k=1}^{n}{k\left( \sum\limits_{r=1}^{r=n-k+1}{r} \right)}=\sum\limits_{k=1}^{n}{k\dfrac{\left( n-k+1 \right)\left( n-k+2 \right)}{2}} \\
& \Rightarrow \sum\limits_{k=1}^{n}{k\left( \sum\limits_{r=1}^{r=n-k+1}{r} \right)}=\sum\limits_{k=1}^{n}{k\dfrac{{{n}^{2}}+{{k}^{2}}-\left( 2n+3 \right)k+3n+2}{2}} \\
& \Rightarrow \sum\limits_{k=1}^{n}{k\left( \sum\limits_{r=1}^{r=n-k+1}{r} \right)}=\sum\limits_{k=1}^{n}{k\dfrac{{{n}^{2}}+3n+2}{2}}-\sum\limits_{k=1}^{n}{k\times k\dfrac{2n+3}{2}}+\dfrac{1}{2}\sum\limits_{k=1}^{n}{k\times {{k}^{2}}} \\
\end{align}\]
We see that the terms involving $n$ are constant and they can be taken outside of the summation. So we have
\[\Rightarrow \sum\limits_{k=1}^{n}{k\left( \sum\limits_{r=1}^{r=n-k+1}{r} \right)}=\dfrac{{{n}^{2}}+3n+2}{2}\sum\limits_{k=1}^{n}{k}-\dfrac{2n+3}{2}\sum\limits_{k=1}^{n}{{{k}^{2}}}+\dfrac{1}{2}\sum\limits_{k=1}^{n}{{{k}^{3}}}\]
We use formulas for the sum of first $n$ terms , the sum of first $n$ squared terms and the sum of first $n$ cubed terms and get,
\[\begin{align}
& =\dfrac{{{n}^{2}}+3n+2}{2}\left( \dfrac{n\left( n+1 \right)}{2} \right)-\dfrac{2n+3}{2}\left( \dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6} \right)+\dfrac{1}{2}{{\left( \dfrac{n\left( n+1 \right)}{2} \right)}^{2}} \\
& =\dfrac{n\left( n+1 \right)}{2}\left( \dfrac{{{n}^{2}}+3n+2}{2}-\dfrac{\left( 2n+3 \right)\left( 2n+1 \right)}{6}+\dfrac{n\left( n+1 \right)}{4} \right) \\
& =\dfrac{n\left( n+1 \right)}{2}\left( \dfrac{6{{n}^{2}}+18n+12}{12}-\dfrac{8{{n}^{2}}+16n+6}{12}+\dfrac{3{{n}^{2}}+3n}{12} \right) \\
& =\dfrac{n\left( n+1 \right)\left( {{n}^{2}}+5n+6 \right)}{24} \\
\end{align}\]
We replace ${{n}^{2}}+5n+6$ with its factorization $\left( n+2 \right)\left( n+3 \right)$ in the above result and get
\[\sum\limits_{k=1}^{n}{k\left( \sum\limits_{r=1}^{r=n-k+1}{r} \right)}=\dfrac{n\left( n+1 \right)\left( n+2 \right)\left( n+3 \right)}{24}\]
Now we can put the above result which is free of $k$ in the given limit and get,
\[\begin{align}
& \displaystyle \lim_{x \to \infty}\dfrac{1}{{{n}^{4}}}\left( 1\sum\limits_{k=1}^{n}{k}+2\sum\limits_{k=1}^{n-1}{k}+3\sum\limits_{k=1}^{n-2}{k...+n\cdot 1} \right)=\displaystyle \lim_{x \to \infty}\dfrac{1}{{{n}^{4}}}\sum\limits_{k=1}^{n}{k\left( \sum\limits_{r=1}^{r=n-k+1}{r} \right)} \\
& =\displaystyle \lim_{x \to \infty}\dfrac{1}{{{n}^{4}}}\left( \dfrac{n\left( n+1 \right)\left( n+2 \right)\left( n+3 \right)}{24} \right) \\
\end{align}\]
Let us take $n$ common from $\left( n+1 \right),\left( n+2 \right),\left( n+3 \right)$ and get
\[\begin{align}
& =\displaystyle \lim_{x \to \infty}\dfrac{1}{{{n}^{4}}}\left( \dfrac{{{n}^{4}}\left( 1+\dfrac{1}{n} \right)\left( 1+\dfrac{2}{n} \right)\left( n+\dfrac{3}{n} \right)}{24} \right) \\
& =\displaystyle \lim_{x \to \infty}\dfrac{\left( 1+\dfrac{1}{n} \right)\left( 1+\dfrac{2}{n} \right)\left( n+\dfrac{3}{n} \right)}{24} \\
\end{align}\]
Now e use the law of multiplication of limits and get
\[\begin{align}
& =\dfrac{\displaystyle \lim_{x \to \infty}\left( 1+\dfrac{1}{n} \right)\displaystyle \lim_{x \to \infty}\left( 1+\dfrac{2}{n} \right)\displaystyle \lim_{x \to \infty}\left( n+\dfrac{3}{n} \right)}{24} \\
& \Rightarrow \displaystyle \lim_{x \to \infty}\dfrac{1}{{{n}^{4}}}\left( 1\sum\limits_{k=1}^{n}{k}+2\sum\limits_{k=1}^{n-1}{k}+3\sum\limits_{k=1}^{n-2}{k...+n\cdot 1} \right)=\dfrac{1\times 1\times 1}{24}=\dfrac{1}{24} \\
\end{align}\]
Note: We know that the limit exists for real valued single variable function $f\left( x \right)$ at any point $x=a$ if and only if Left hand limit$\left( \displaystyle \lim_{x \to {{a}^{-}}}f\left( x \right) \right)=$right hand limit$\left( \displaystyle \lim_{x \to {{a}^{+}}}f\left( x \right) \right)$. We note that we have only evaluated only right hand limit because the function is continuous everywhere in $R.$
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