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<p><img src="https://sites.google.com/site/calculusproject13/_/rsrc/1472871578846/home/INFINITE%20SERIES%20HANDOUT.jpg" width="991" height="1083"/></p>
<p> Hello it's me again drifter1! Today we continue with <strong>Mathematical Analysis</strong> getting into more about <strong>Series</strong>. After talking about the Basics in my last post <a href="https://steemit.com/mathematics/@drifter1/mathematics-mathematical-analysis-series-basics">here</a>, we will now get into how we<strong> test the Convergence</strong>. So, without further do, let's get started!</p>
<h2>Cauchy's Convergence Criterion:</h2>
<p>A Series s converges only when for every e>0 there is a natural number n0 such that: </p>
<p>|S| < e => |a(n) + a(n+1) + ... + a(n+m)| < e for every m, n with m>=n>n0</p>
<p>We can also describe it as:</p>
<ul>
<li>for every e>0 there is a natural n0 such that |Sm - Sn| < e for every m, n with m>=n>n0</li>
<li>for every e>0 there are natural p, n0<=n such that |S(n+p) - Sn| < e <=> lim n->∞ (S(n+p) - Sn) = 0</li>
</ul>
<p><strong>Example proof:</strong></p>
<p>Using this Criterion we can easily proof that the harmonic series with p=1 (1/n terms) diverges.</p>
<p>For m = 2n there is a n0 so that for every e>0 and n>=n0: |S2n - Sn| < e</p>
<p>The generic term of such a series is a(n) = 1/n and so:</p>
<p>S2n - Sn = 1/n+1 + 2/n+2 + ... + n/2n > n/2n = 1/2 => |S2n - Sn| >= 1/2.</p>
<p>And so Cauchy's criterion is not true and the series diverges.</p>
<p><br></p>
<p>So, where does this help us?</p>
<p>Well, think about the <strong>limit of (an) as a testing method of convergence</strong>.</p>
<ul>
<li>When the series converges then the lim n->∞ (an) = 0</li>
<li>If lim n->∞(an) !=0 then the series diverges</li>
</ul>
<p><em><strong>ATTENTION:</strong></em><em> </em><strong>It doesn't mean that a sequence converges when the limit is 0</strong>, but <strong>if it isn't 0 then we are 100% sure that the series diverges</strong>. Think about the harmonic series from before, where the limit of 1/n to infinity is equal to 0, but the series diverges!</p>
<p><br></p>
<h2>Sequence Boundary Test:</h2>
<p>Suppose s is positive series (an >=0 for every natural n).</p>
<ol>
<li>s converges only if the sum of n-terms (sn = a1 + a2 + ...+ an) is bounded</li>
<li>s diverges if the sum of n-terms (sn = a1 + a2 + ...+ an) is not-bounded</li>
</ol>
<h2><br></h2>
<h2>Comparison Test:</h2>
<p>Suppose s1 and s2 are two positive series with terms an and bn and 0 <= an <= bn for every natural n.</p>
<ul>
<li>If s2 converges then s1 converges also</li>
<li>If s2 diverges to ∞ then s1 also diverges to ∞</li>
</ul>
<p> This means that we can check the convergence/divergence of a "greater" series to check the convergence/divergence of a "smaller" series.</p>
<p><br></p>
<h2>Series Addition Convergence:</h2>
<p> Suppose the Series s1, s2 with terms an, bn and the real numbers k, l. If the series converge to a and b respectively then the series S with term k*an + l*bn converges to k*a + l*b.</p>
<p>Cases:</p>
<ul>
<li>If both converge then the result series S also converges</li>
<li>If at least one diverges then the result series S also diverges</li>
</ul>
<p>So,<strong> in a sum of Series,</strong> <strong>if at least one diverges then the whole result also diverges</strong>!</p>
<p><br></p>
<h2>Condensation Test:</h2>
<p>Suppose (an) is a decreasing positive sequence (an>=an+1>=0).</p>
<ol>
<li>The <strong>series with term (an) converges</strong> <strong>only if the series with term 2^k*a(2^k) converges</strong> and vise versa</li>
<li>In the same way <strong>if one diverges then the other diverges too</strong></li>
</ol>
<p><br></p>
<p><strong>Application:</strong></p>
<p>Let's talk about <strong>harmonic series of p-degree</strong> with p being a real number and terms 1/n^p.</p>
<ul>
<li>The series <strong>converges when p>1</strong></li>
<li>The series <strong>diverges to +∞ when p<=1</strong></li>
</ul>
<p>This can be proven easily using the condensation test.</p>
<p>For p<=0 we get a series with terms n^k, with k =-p and so the series diverges to +∞.</p>
<p>For p>0 and using the condensation test we check the convergence of the series with term:</p>
<p>2^k/(2^k)^p = 2^(1-p)^k = x^k, where x^k is a geometric series and so we have that:</p>
<p>The series converges when: 2^(1-p) < 1 = 2^0 => p > 1 </p>
<p>The series diverges when: 2^(1-p) >= 1 = 2^0 => p <= 1</p>
<p> We know that the first series also converges/diverges for the same values, because of the condensation convergence test.</p>
<p><br></p>
<h2>Limit comparison test:</h2>
<p>Suppose the series s1, s2 with positive terms an and bn and the limit n->∞ (an/bn) = k.</p>
<ol>
<li>If 0 < k < + ∞ then s1 and s2 converge/diverge together</li>
<li>If k = 0 then if s2 converges then s1 also converges (NOT vise versa)</li>
<li>If k = +∞ then if s1 converges then s2 also converges (NOT vise versa)</li>
</ol>
<p>This is pretty useful in series build up of polynomial sequences.</p>
<p><br></p>
<h2>Ratio test:</h2>
<p>Suppose s is a non-zero series with term (an) and lim n->∞|a(n+1)/an| = l, with l > 0.</p>
<ol>
<li>If l < 1 then the series converges</li>
<li>If l > 1 then the series diverges to +∞ and lim n->∞ (an) != 0</li>
<li>If l =1 then we don't know and need to use some other test.</li>
</ol>
<p>This one is pretty useful when having a rational term build up of exponentials, factorials and more.</p>
<h2>n-root Test:</h2>
<p>Suppose s a series with term (an) and lim n->∞ [n-root(|an|)] = l, with l > 0.</p>
<ol>
<li>If 0 <= l < 1 then the series converges</li>
<li>If l > 1 then the series diverges to +∞</li>
<li>If l = 1 then we don't know and need to use some other test.</li>
</ol>
<p>This one is pretty useful in series build up of polynomial sequences to get rid of high exponentials.</p>
<p><br></p>
<h2>Absolute test:</h2>
<p>When s is a Series build up of terms (an) and |s| is a sereis build up of terms |(an)| then:</p>
<ul>
<li>If |s| converges then s also converges, but the opposite is not true!</li>
<li>We don't know about divergence</li>
</ul>
<p><br></p>
<h2>Integral test:</h2>
<p>Suppose a integratable, positive and increasing function f:[1, +∞) -> R.</p>
<p>The definite integral:</p>
<p><img src="https://s8.postimg.org/6e5al0p6d/Screenshot_1.jpg" width="98" height="57"/></p>
<p>and the Series:</p>
<p><img src="https://s8.postimg.org/9kzu4nhc5/Screenshot_2.jpg" width="99" height="52"/></p>
<p>converge/diverge together.</p>
<p>When their converge then we know that <em>I < s < I + f(1)</em> with I:</p>
<p><img src="https://s8.postimg.org/c2blby3th/Screenshot_3.jpg" width="209" height="60"/></p>
<p><br></p>
<h2>Leibniz Criterion for Alternating Series:</h2>
<p> <strong>When the absolute value of the terms of an alternating series are a decreasing and null-sequence then the series converges</strong>.</p>
<p>So, for an alternating series with term (-1)^(n+1) * (an) we have that <strong>it converges when</strong>:</p>
<ul>
<li><strong>a(n) >= a(n+1) >= 0</strong> for every natural n</li>
<li><strong>lim n->∞ (an) = 0</strong></li>
</ul>
<p><br></p>
<p><strong>Sum/Limit approximation:</strong></p>
<p>When such a series converges then we can calculate the limit approximately.</p>
<p>The sum of n-terms is: a1 - a2 + a3 - ... + (-1)^(n+1)*(an).</p>
<p>This value comes close to the sum of our series s having an error.</p>
<p>The value of this error is smaller then the absolute value of the (n+1)-term: |s - Sn| <= an+1</p>
<p>The remainder (s-Sn) has the same sign as the (n+1)-term.</p>
<p><br></p>
<p>And this is actually it for today and I hope you learned something!</p>
<p> The post would become too clumped up if I also put Examples for everything today. That's why we will get into examples for all (or some) of the convergence tests we covered today in my next post. </p>
<p>Until next time...Bye!</p>
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