It's actually a much simpler equation than the one for the first n terms, but it only works if -1< r<1Įxample 1: If the first term of an infinite geometric series is 4, and the common ratio is 1/2, what is the sum?Įxample 2: The sum of an infinite geometric series is 36, and the common ratio is 1/3. And just like that, we have the equation for S, the sum of an infinite geometric series: In fact, we could say that when n goes to infinity, r ngoes to zero. It gets really, really close to zero, doesn't it? For faster calculations, use our calculator by just giving the inputs in the input fields and getting the concerned output. Then what happens to r n when n gets really big? An infinite sequence calculator is an online tool that helps to calculate the sum of the given function for the given limits. Suppose r was less than 1, but greater than -1. Let me show you.įrom the previous page in this unit, we know that The feeling we have about numerical methods like Newton'smethod and the bisection method is that if we continue theiteration process more and more times, we would get numbers thatare closer and closer to the actual root of the equation. However whenever I run the code what I am returned is the value of x or the same value I input. But for some series it is possible to find the sum of an infinite number of terms, and even though that might seem like a lot of work, it's really pretty simple. Write C++ function to evaluate the following formula for a given x: The following code was designed in C++ on Visual Studio to be a solution of the above mentioned problem. If $\alpha + \beta = 1$ you basically pull the same trick starting at $n = 2$ instead, with $F(1) = 0$.Up until now we've only looked at the sum of the first n terms of a geometric series (S n). You choose $F(n)$ in increasing order of $n$ by choosing $F(n)$ to be anything you want for all $n \leq N$, and then for $F(n)$ for $n > N$ you choose $F(n)$ so that the partial sum going up to the $n$th term in your series, call is $S_n$, satisfies $|S_n| \leq c |S_^\infty 1/2^n = -1$, so your series is zero. If so, you should update your question to reflect that (since there are iterative procedures that will create uncountably many solutions, even up to scaling, although they probably won't provide much insight as to how to get elementary closed form solutions.)Īlso, if you are interested in an iterative procedure to construct $F(n)$, here's the basic idea. Perhaps that is the thrust of your question, how to find an elementary closed form solution. You can construct such $F(n)$ iteratively that satisfy your infinite series equation, making arbitrary choices of certain constants along the way, but I'm not sure whether there is an elementary closed form formula for $F(n)$ that works. So one of the only things you can really say at first glance is that $F(n)$ tends to zero fast enough so that your series expression terms are able to, i.e. Begin by finding the common ratio, r 6 3 2. Example 9.3.1: Find an equation for the general term of the given geometric sequence and use it to calculate its 10th term: 3, 6, 12, 24, 48. For example, all you need for $F(n)$ is that your sequence $F(n)(1-(\alpha n + \beta)^2)$ tends to $0$ in the "right way" to get the series formula to converge to zero, and you're done. In fact, any general term that is exponential in n is a geometric sequence. But that doesn't even begin to describe the rich variety you can have. I'm not sure if this helps, but there are (uncountably) infinitely many such $F(n)$, for multiple reasons, one being that if $F(n)$ is a solution then $cF(n)$ is also a solution for any constant $c$.
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