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For the series $$\sum_{k=0}^{n}x^k=\frac{1-x^{n+1}}{1-x}$$if I take the limit as $n\rightarrow \infty$, I am getting confused regarding the limit for the right hand side, whether, it will be $-\infty$ or else.$$ \sum_{k=0}^{\infty}x^k=\lim_{n \to \infty }\frac{1-x^{n+1}}{1-x}$$

PinkyWay
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Roshan Shrestha
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    The series is convergent only for $|x|<1$ (for $|x|\ge 1$ the general term does not have limit $0$). In this case the limit is just $\frac 1{1-x}$. – zwim Apr 28 '20 at 11:04
  • Thanks, but is there any way we can show it. – Roshan Shrestha Apr 28 '20 at 11:07
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    I guess you can show it using the fact that $x^n$ as $n$ tends to infinity equals $0$ if and only if $|x|<1$ – UnsinkableSam Apr 28 '20 at 11:10
  • I can do it roughly taking x to be between 0 and 1 and show that it converges only for $\left | x \right | < 1$, but is there any test we can do to show that the series converges only for this region. – Roshan Shrestha Apr 28 '20 at 11:17
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    I mean this series diverge if $|x|>1$ by Ratio Test so I guess what you need to show is the case of $x=1$ and $x=-1$ but that should be relatively obvious. I hope I answered some question here – UnsinkableSam Apr 28 '20 at 11:22

3 Answers3

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Using that $x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$ (you can find that proved in this link Proving $x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$), you can replace the expression with $$\lim_{n \to \infty }(1^{n-1} + 1^{n-2} *x + ... + 1* x^{n-2} + x^{n-1}),$$ what equals $$\lim_{n \to \infty }(x + ... + x^{n-2} + x^{n-1}).$$ And obviously $$\lim_{n \to \infty }(x + ... + x^{n-2} + x^{n-1})= \lim_{n \to \infty }\sum_{k=0}^{n}x^k =\sum_{k=0}^{\infty}x^k.$$

Alejandro Bergasa Alonso
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$$ \lim_{n \to \infty} \frac{1-x^{n+1}}{1-x}=\infty,~~ if~~ |x|>1$$ $$ \lim_{n \to \infty} \frac{1-x^{n+1}}{1-x}=\frac{1}{1-x},~~ if~~ |x|<1.$$

Z Ahmed
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The only part you need consider on RHS is the term that depends on $n,$ namely $x^{n+1}.$

So let's consider the limit of $x^m$ where $m$ is a nonnegative integer. This is simply the $m$-fold product $$\underbrace{xxxxxx\cdots x}_{m}.$$

Now if you increase the number of factors in a product like this, the product will behave differently according as $|x|=1,\,|x|<1$ and $|x|>1.$ Thus, to understand this limit as the number of factors becomes infinite, you need to understand the nature of self-multiplication. It suffices to understand the nature of $yy=y^2$ for $y\ge 0.$ Now, if $y<1,$ then multiplying both sides by $y>0$ gives $$y^2<y<1.$$ Thus, multiplying a small number (that is, a positive number less than $1$) by itself reduces the product. If $y>1,$ you can similarly see that the product increases. If $y=1,$ of course nothing happens.

Now come back to your product. If the terms are absolutely bigger than $1$ then the sum increases without bound in absolute value. In this case the sum has no real value. On the other hand if $|x|<1,$ then we have seen that the product becomes ever smaller, which means that it approaches $0$ arbitrarily closely as the number of factors becomes infinite. Finally, if $x=-1,$ you can see that the sequence $(-1)^n$ oscillates, thus never converges, so that the sum does not converge in this case too; and when $x=1,$ the expression on RHS is invalid. However, in this case the series is just $1+1+1+1+\cdots,$ which clearly converges to no real value.

In summary, your sum converges to the real number $$\frac{1}{1-x}$$ as the summands become infinitely many, when and only when $|x|<1.$

Allawonder
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